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KIRBY ELEMENTS AND QUANTUM INVARIANTS

ALEXIS VIRELIZIER

Abstract. We define the notion of a Kirby element of a ribbon category C (notnecessarily semisimple). Kirby elements lead to 3-manifolds invariants. Wecharacterize a set of Kirby elements (in terms of the structural morphisms of aHopf algebra in C) which is sufficiently large to recover the quantum invariantsof 3-manifolds of Reshetikhin-Turaev, of Hennings-Kauffman-Radford, and ofLyubashenko. The cases of a semisimple ribbon category and of a category ofrepresentations are explored in details.

Contents

Introduction 11. Ribbon categories and coends 32. Kirby elements of a ribbon category 93. The case of semisimple ribbon categories 154. The case of categories of representations 255. A non-unimodular example 41Appendix A. Traces on ribbon Hopf algebras 43References 44

Introduction

During the last decade, deep connections between low-dimensional topology andthe purely algebraic theory of quantum groups (or, more generally, of braided cat-egories) were highlighted. In particular, this led to a new class of 3-manifoldsinvariants, called quantum invariants, defined in several ways.

The aim of the present paper is to give a as general as possible method ofconstructing quantum invariants of 3-manifolds starting from a ribbon category ora ribbon Hopf algebra. With this formalism, we recover the 3-manifolds invariants ofReshetikhin-Turaev [RT91, Tur94], of Hennings-Kauffman-Radford [KR95, Hen96],and of Lyubashenko [Lyu95a] when these are well-defined.

Let k be a field and C be a k-linear ribbon category (not necessarily semisimple).Under some technical assumption, namely the existence of a coend A ∈ Ob(C) ofthe functor (X,Y ) ∈ Cop ×C 7→ X∗ ⊗ Y ∈ C, a scalar τC(L;α) can be associated toany framed link L in S3 and any morphism α ∈ HomC(1, A), see [Lyu95a]. Recall,see [Lyu95b], that the object A of C is then a Hopf algebra in the category C.

By a Kirby element of C, we shall mean a morphism α ∈ HomC(1, A) suchthat τC(L;α) is invariant under isotopies of L and under 2-handle slides. By using

Date: November 13, 2018.2000 Mathematics Subject Classification. 57M27,18D10,81R50.

1

http://arxiv.org/abs/math/0312337v2

2 ALEXIS VIRELIZIER

the Kirby theorem [Kir78], we have that if α is a Kirby element of C such thatτC(©

±1;α) 6= 0, then τC(L;α) can be normalized to an invariant τC(ML;α) of3-manifolds, where ©±1 is the unknot with framing ±1 and ML denotes the 3-manifold obtained from S3 by surgery along L.

In general, determining all the Kirby elements of C is a quite difficult problem.In this paper we characterize, in terms of the structure maps of the categoricalHopf algebra A, a set I(C) made of Kirby elements of C which is sufficiently largeto contain the Kirby elements corresponding to the known quantum invariants.

If the categorical Hopf algebra A admits a two-sided integral λ : 1 → A, thenλ belongs to I(C) and the corresponding invariant τC(M ;λ) is the Lyubashenkoinvariant [Lyu95a].

When C is semisimple, we give necessary conditions for being in I(C) and we showthat there exist (even in the non-modular case) elements of I(C) corresponding tothe Reshetikhin-Turaev invariants [RT91, Tur94] computed from finitely semisimplefull ribbon subcategories of C. Note that these elements are not in general two-sidedintegrals.

More generally, we show that I(C) contains the Kirby elements leading to theinvariants obtained from I(B), where B is a finitely semisimple full ribbon subcat-egory of the semisimple quotient of C. We verify that, in general, there exist Kirbyelements in I(C) which are not of this last form. This means that the semisimpli-fication process “lacks” some invariants.

Let H be a finite-dimensional ribbon Hopf algebra. Suppose that C is the cat-egory repH of finite-dimensional left H-modules. We describe I(H) = I(repH) inpurely algebraic terms. One of the interest of such a description is to avoid therepresentation theory of H (which may be of wild type, see [Ben98]).

If H is unimodular, then 1 ∈ I(H) and the corresponding invariant τ(H,1) is theHennings-Kauffman-Radford invariant [KR95, Hen96]. More generally, and even ifH is not unimodular, we show that the invariant τ(H,z) of 3-manifolds correspondingto z ∈ I(H) can be computed by using the Kauffman-Radford algorithm.

If V is a set of simple left H-modules which makes (H,V) a premodular Hopfalgebra, then there exists zV ∈ I(H) such that τ(H,zV) is the Reshetikhin-Turaev in-variant computed from (H,V), which can then be computed by using the Kauffman-Radford algorithm.

When H is semisimple and k is of characteristic 0, we show that the Hennings-Kauffman-Radford invariant (computed from H) and the Reshetikhin-Turaev in-variant (computed from repH) are simultaneously well-defined and coincide (evenin the non-modular case). In the modular case, this was first shown in [Ker97].

We explicitly determine I(H) for a family of non-unimodular ribbon Hopf alge-bras which contains Sweedler’s Hopf algebra.

As an algebraic application, the operators involved in the description of I(H) =I(repH) in algebraic terms allows us to parameterize all the traces on a finite-dimensional ribbon Hopf algebra H . When H is unimodular, we recover the pa-rameterization given in [Rad94b, Hen96].

The paper is organized as follows. In Section 1, we review ribbon categoriesand coends. In Section 2, we define and study Kirby elements. We focus, in Sec-tion 3, on the case of semisimple ribbon categories and, in Section 4, on the caseof categories of representations of ribbon Hopf algebras. In Section 5, we treat an

KIRBY ELEMENTS AND QUANTUM INVARIANTS 3

example in detail. Finally, in Appendix A, we study traces on ribbon Hopf algebras.

Acknowledgements. I thank A. Bruguieres for helpful discussions, in particularfor questions concerning category theory.

1. Ribbon categories and coends

In this section, we review some basic definitions concerning ribbon categoriesand coends. Throughout this paper, we let k be a field.

1.1. Ribbon categories. Let C be a strict monoidal category with unit object 1(note that every monoidal category is equivalent to a strict monoidal category in acanonical way, see, [Lan98]). A left duality in C associates to any object U ∈ C anobject U∗ ∈ C and two morphisms evU : U∗⊗U → 1 and coevU : 1→ U ⊗U∗ suchthat

(idU ⊗ evU )(coevU ⊗ idU ) = idU(1.1)

(evU ⊗ idU∗)(idU∗ ⊗ coevU ) = idU∗ .(1.2)

We can (and we always do) impose that 1∗ = 1, ev1

= id1

and coev1

= id1

.By a braided category we shall mean a monoidal category C with left duality and

endowed with a system {cU,V : U ⊗V → V ⊗U}U,V∈C of invertible morphisms (thebraiding) satisfying the following three conditions:

cU ′,V ′(f ⊗ g) = (g ⊗ f) cU,V ,(1.3)

cU⊗V,W = (cU,W ⊗ idV )(idU ⊗ cV,W ),(1.4)

cU,V ⊗W = (idV ⊗ cU,W )(cU,V ⊗ idW ),(1.5)

for all objects U, V,W ∈ C and all morphisms f : U → U ′, g : V → V ′ in C. Notethat, by applying (1.4) to U = V = 1 and (1.5) to V = W = 1 and using theinvertibility of cU,1 and c

1,U , we obtain that cU,1 = c1,U = idU for any object

U ∈ C.A ribbon category is a braided category C endowed with a family of invertible

morphisms {θU : U → U}U∈C (the twist) satisfying the following conditions:

θV f = f θU ;(1.6)

(θU ⊗ idU∗)coevU = (idU ⊗ θU∗)coevU ;(1.7)

θU⊗V = cV,U cU,V (θU ⊗ θV );(1.8)

for all objects U, V ∈ C and all morphism f : U → V in C. It follows from (1.8)that θ

1

= id1

.A ribbon category C canonically has a right duality by associating to any object

U ∈ C its left dual U∗ ∈ C and two morphisms evU : U ⊗ U∗ → 1 and coevU : 1→U∗ ⊗ U defined by

evU = evU cU,U∗(θU ⊗ idU∗);(1.9)

coevU = (idU∗ ⊗ θ−1U )(cU∗,U )

−1coevU .(1.10)

Note that we have ev1

= id1

and coev1

= id1

.

4 ALEXIS VIRELIZIER

The dual morphism f∗ : V ∗ → U∗ of a morphism f : U → V in a ribbon categoryC is defined by

f∗ = (evV ⊗ idU∗)(idV ∗ ⊗ f ⊗ idU∗)(idV ∗ ⊗ coevU )(1.11)

= (idU∗ ⊗ evV )(idU∗ ⊗ f ⊗ idV ∗)(coevU ⊗ idV ∗).

It is well-known that (idU )∗ = idU∗ and (fg)∗ = g∗f∗ for composable morphisms

f, g. Axiom (1.7) can be shown to be equivalent to θ∗U = θU∗ .Let C be a ribbon category. Note that EndC(1) is a semigroup, with composition

as multiplication, which is commutative (since C is monoidal). The quantum traceof an endomorphism f : U → U of an object U ∈ C is defined by

(1.12) trq(f) = evU (f ⊗ idU∗) coevU = evU (idU∗ ⊗ f) coevU ∈ EndC(1).

For any morphisms u : U → V , v : V → U and for any endomorphisms f, g in C,we have

(1.13) trq(uv) = trq(vu), trq(f∗) = trq(f), and trq(f ⊗ g) = trq(f) trq(g).

The quantum dimension of an object U ∈ C is defined by

(1.14) dimq(U) = trq(idU ) = evU coevU = evU coevU ∈ EndC(1).

Isomorphic objects have equal dimensions and dimq(U ⊗ V ) = dimq(U) dimq(V )for any objects U, V ∈ C. Note that dimq(1) = id

1

.

1.2. k-categories. Let k be a field. By a k-category, we shall mean a categoryfor which the sets of morphisms are k-spaces and the composition is k-bilinear.By a monoidal k-category, we shall mean k-category endowed with a monoidalstructure whose tensor product is k-bilinear. Note that if C is a monoidal k-category,then EndC(1) is a commutative k-algebra (with composition as multiplication). Amonoidal k-category is said to be pure if EndC(1) = k.

By a ribbon k-category, we shall mean a pure monoidal k-category endowed witha ribbon structure.

1.3. Graphical calculus. Let C be ribbon category. Any morphism in C can begraphically represented by a plane diagram (we use the conventions of [Tur94]).This pictorial calculus will allow us to replace algebraic arguments involving com-mutative diagrams by simple geometric reasoning. This is justified, e.g., in [Tur94].

A morphism f : V → W in C is represented by a box with two vertical arrowsoriented downwards, as in Figure 1(a). Here V,W should be regarded as “colors”of the arrows and f should be regarded as a “color” of the box. More generally, amorphism f : V1⊗· · ·⊗Vm →W1⊗· · ·⊗Wn may be represented as in Figure 1(b).

We also use vertical arrows oriented upwards under the convention that themorphism sitting in a box attached to such an arrow involves not the color of thearrow but rather the dual object.

The identity endomorphism of an object V ∈ C or of its dual V ∗ will be repre-sented by a vertical arrow as depicted in Figure 1(c). Note that a vertical arrowcolored with 1 may be deleted from any picture without changing the morphismrepresented by this picture. The symbol “=” displayed in the figures denotes equal-ity of the corresponding morphisms in C.

The tensor product f⊗g of two morphisms f and g in C is represented by placinga picture of f to the left of a picture of g. A picture for the composition g ◦ f of


PSfrag replacementsf

V

W

(a) f : V → W

...

...

PSfrag replacements

f

V1

W1

Vm

Wn

(b) Tensor product

PSfrag replacements

idV ∗

V

V

idV

V ∗

= ==

(c) The identity

Figure 1. Plane diagrams of morphisms

two (composable) morphisms g and f is obtained by putting a picture of g on thetop of a picture of f and by gluing the corresponding free ends of arrows.

The braiding cV,W : V ⊗W → W ⊗ V and its inverse c−1V,W : W ⊗ V → V ⊗W ,

the twist θV : V → V and its inverse θ−1V : V → V , and the duality morphisms

evV : V ∗⊗V → 1, coevV : 1 → V ⊗V ∗, evV : V ⊗V ∗ → 1, and coevV : 1→ V ∗⊗Vare represented as in Figures 2(a), 2(b), and 2(c) respectively. The quantum traceof an endomorphism f : V → V in C and the quantum dimension of an objectV ∈ C may be depicted as in Figure 2(d).

1.4. Negligible morphisms. Let C be a ribbon k-category. A morphism f ∈HomC(X,Y ) is said to be negligible if trq(gf) = 0 for all g ∈ HomC(Y,X). Denoteby NeglC(X,Y ) the k-subspace of HomC(X,Y ) formed by the negligible morphisms.

It is important to note that NeglC is a two-sided ⊗-ideal of C. This means thatthe composition (resp. the tensor product) of a negligible morphism with any othermorphism is negligible.

Note that since End(1) = k, a morphism f : 1 → X is negligible if and only ifgf = 0 for all morphism g : X → 1.

1.5. Dinatural transformations and coends. Recall that to each category Cwe associate the opposite category Cop in the following way: the objects of Cop

are the objects of C and the morphisms of Cop are morphisms fop, in one-onecorrespondence f 7→ fop with the morphisms in C. For each morphism f : U → Vof C, the domain and codomain of the corresponding fop are as in fop : V → U(the direction is reversed). The composite fopgop = (gf)op is defined in Cop exactlywhen the composite gf is defined in C. This makes Cop a category.

Let C and B be two categories. A dinatural transformation between a functorF : Cop × C → B and an object B ∈ B is a function d which assigns to each objectX ∈ C a morphism dX : F (X,X) → B of B in such a way that the diagram

F (Y,X)F (idY ,f)−−−−−−→ F (Y, Y )

F (f,idX)

yydY

F (X,X)dX−−−−→ B

commutes for every morphism f : X → Y in C.A coend of the functor F is a pair (A, i) consisting of an object A of B and a

dinatural transformation i from F to A which is universal among the dinaturaltransformation from F to a constant, that is, with the property that, to every

6 ALEXIS VIRELIZIERPSfrag replacements

cV,W

VV

V V

VV WW

WW

W

W

W

W

Wc−1V,W

==

(a) Braiding

PSfrag replacements

θ−1V

VV

VV VV

VV

θV = ===

(b) Twist

PSfrag replacements

evVVV

VVevV coevVcoevV = ===

(c) Duality morphismsPSfrag replacements

V VdimqVtrq(f) f= =

(d) Trace and dimension

Figure 2.

dinatural transformation d from F to B, there exists a unique morphism r : A→ Bsuch that, for all object X ∈ C,

(1.15) dX = r ◦ iX .

By using the factorization property (1.15), it is easy to verify that if (A, i) and(A′, i′) are two coends of F , then they are isomorphic in the sense that there existsan isomorphism I : A→ A′ in B such that i′X = I ◦ iX for all object X ∈ C.

1.6. Categorical Hopf algebras from coends. Let C be a ribbon category.Consider the functor F : Cop × C → C defined by

(1.16) F (X,Y ) = X∗ ⊗ Y and F (fop, g) = f∗ ⊗ g

for all objects X ∈ Cop, Y ∈ C and all morphisms f, g in C.Suppose that the functor F admits a coend (A, i). Then the object A has a

structure of a Hopf algebra in the category C (see [Lyu95b]). This means that thereexist morphisms mA : A ⊗ A → 1, ηA : 1 → A, ∆A : A → A ⊗ A, εA : A → 1,and SA : A → A, which verify the same axioms as those of a Hopf algebra except


the usual flip is replaced by the braiding cA,A : A ⊗ A → A ⊗ A. By using thefactorization property (1.15) and the naturality of the braiding and twist of C,these structural morphisms are defined as follows:

∆A ◦ iX =(X∗ ⊗X

idX∗⊗coevX⊗idX✲ X∗ ⊗X ⊗X∗ ⊗X

iX⊗iX✲ A⊗A

),(1.17)

εA ◦ iX = evX ,(1.18)

ηA : 1 = 1

∗ ⊗ 1

i1

✲ A,(1.19)

mA ◦ (iX ⊗ iY ) =(X∗ ⊗X ⊗ Y ∗ ⊗ Y

cX,Y ∗⊗Y✲ X∗ ⊗ Y ∗ ⊗ Y ⊗X(1.20)

≃ (Y ⊗X)∗ ⊗ (Y ⊗X)iY ⊗X

✲ A),

SA ◦ iX =(X∗ ⊗X

cX∗,X✲ X ⊗X∗ idX⊗θX∗

✲ X ⊗X∗(1.21)

≃ X∗∗ ⊗X∗ iX∗✲ A

).

Here X and Y are any objects of C. The defining relations (1.17)-(1.21) are graph-ically represented in Figure 3.

PSfrag replacements

A

A

AA A

X XX

X

X

iX

iXiX ∆A

=

(a) ∆A : A → A⊗A

PSfrag replacementsA

X XXX

iX

εA

=

(b) εA : A → 1

PSfrag replacements

A AAX

i1

ηA

=

(c) ηA : 1 → A

PSfrag replacements

A

AA

A

idY⊗X

idY⊗X

XXXX YYYY

Y

=

Y

Y ⊗X

Y ⊗X

iY⊗X

iX iY

mA

(d) mA : A⊗A → A

PSfrag replacements A

A

A

idXidX

iX∗

iX

SA

XXXX

XX

X∗

=

X∗

(e) SA : A → A

Figure 3. Structural morphisms of A

8 ALEXIS VIRELIZIER

It can be shown that the antipode SA : A → A is an isomorphism and thatS2A = θA (see [Lyu95b]). Moreover, as for Hopf algebras, the antipode is anti-

(co)multiplicative, that is,

SAmA = mA(SA ⊗ SA)cA,A, SAηA = ηA,(1.22)

∆ASA = cA,A(SA ⊗ SA)∆A, εASA = εA.(1.23)

The Hopf algebra A posses a Hopf pairing ωA : A⊗A→ 1 (see [Lyu95a]) definedby

(1.24) ωA ◦ (iX ⊗ iY ) = (evX ⊗ evY )(idX∗ ⊗ cY ∗,XcX,Y ∗ ⊗ idY )

for any objects X,Y ∈ C. This pairing is said to be non-degenerate if ωA(idA ⊗coevA) : A→ A∗ and ωA(coevA ⊗ idA) : A→ A∗ are isomorphisms.

Set:

Γl = (idA ⊗mA)(∆A ⊗ idA) : A⊗A→ A⊗A,(1.25)

Γr = (mA ⊗ idA)(idA ⊗∆A) : A⊗A→ A⊗A.(1.26)

Lemma 1.1. Γl(SA ⊗ SA)cA,A = cA,A(SA ⊗ SA)Γr.

Proof. By using (1.3), (1.22) and (1.23), we have

Γl(SA ⊗ SA)cA,A = (idA ⊗mA)(∆ASA ⊗ SA)cA,A

= (idA ⊗mA)(cA,A(SA ⊗ SA)∆A ⊗ SA)cA,A

= (SA ⊗mA(SA ⊗ SA))(cA,A∆A ⊗ idA)cA,A

= (SA ⊗ SAmA)(idA ⊗ c−1A,A)(cA,A ⊗ idA)(∆A ⊗ idA)cA,A.

Then, using (1.3) and (1.4)-(1.5), we get that

Γl(SA ⊗ SA)cA,A

= (SA ⊗ SAmA)(idA ⊗ c−1A,A)(cA,A ⊗ idA)cA,A⊗A(idA ⊗∆A)

= (SA ⊗ SAmA)(idA ⊗ c−1A,A)cA,A⊗A(idA ⊗ cA,A)(idA ⊗∆A)

= (SA ⊗ SAmA)(idA ⊗ c−1A,A)(idA ⊗ cA,A)(cA,A ⊗ idA)(idA ⊗ cA,A)(idA ⊗∆A)

= (SA ⊗ SAmA)cA⊗A,A(idA ⊗∆A)

= cA,A(SA ⊗ SA)(mA ⊗ idA)(idA ⊗∆A)

= cA,A(SA ⊗ SA)Γr.

�

Corrolary 1.2. Suppose that C is moreover a (monoidal) k-category. Let α ∈HomC(1, A). If SAα−α ∈ NeglC(1, A), then the following assertions are equivalent:

(a) Γl(α⊗ α)− α⊗ α : 1→ A⊗A is negligible;(b) Γr(α⊗ α)− α⊗ α : 1→ A⊗A is negligible.

Moreover, if SA ◦ α = α, then Γl(α⊗ α) = α⊗α if and only if Γr(α⊗ α) = α⊗ α.

Proof. This is an immediate consequence of Lemma 1.1 since NeglC is a two-sided⊗-ideal of C. �


2. Kirby elements of a ribbon category

In this section, we generalize the Lyubashenko’s method [Lyu95a] of constructing3-manifolds invariants from ribbon categories.

2.1. Special tangles. Let n be a positive integer. By a n-special tangle we shallmean an oriented framed tangle T ⊂ R

2 × [0, 1] with 2n bottom endpoints and notop endpoints, consisting of n arc components such that the kth arc (1 ≤ k ≤ n)joins the (2k − 1)th and (2k)th bottom endpoints and is oriented out of R2 × [0, 1]near the (2k)th bottom endpoint (and so inside R

2 × [0, 1] near the (2k − 1)thbottom endpoint).

Diagrams of special tangles are drawn with blackboard framing. An example ofa 3-special tangle is depicted in Figure 4(a).

Let C be a ribbon category. Suppose that the functor (1.16) admits a coend(A, i). Let T be a n-special tangle. For objets X1, . . . , Xn ∈ C, let T(X1,··· ,Xn) bethe morphism X∗

1 ⊗ X1 ⊗ · · · ⊗ X∗n ⊗ Xn → 1 in C graphically represented by a

diagram of T where the kth component of T has been colored with the object Xk.Note that T(X1,··· ,Xn) does not depend on the choice of the diagram of T , that is,only depends on the isotopy class of T (see [Tur94]). Since the braiding and twistof C are natural and by using the Fubini theorem for coends (see [Lan98]), thereexists a (unique) morphism φT : A⊗n → 1 such that

(2.1) T(X1,...,Xn) = φT ◦ (iX1 ⊗ · · · ⊗ iXn)

for all objects X1, . . . , Xn ∈ C (see Figure 4(b) for n = 3).

(a) A 3-special tangle T

PSfrag replacements

iX1 iX2 iX3

X1X1 X1X1

AAA

X2 X2 X2 X2X3X3 X3X3

T(X1,X2,X3)

φT

=

(b) T(X1,X2,X3) = φT ◦ (iX1⊗ iX2

⊗ iX2)

Figure 4.

2.2. Kirby elements. Let C be a ribbon category such that the functor (1.16)admits a coend (A, i).

Let L be a framed link in S3 with n components. Fix an orientation for L.There always exists a (non-unique) n-special tangle TL such that L is isotopicTL ◦ (∪− ⊗ · · · ⊗ ∪−), where ∪− denote the cup with clockwise orientation, seeFigure 6(a). For α ∈ HomC(1, A), set

τC(L;α) = φTL◦ α⊗n ∈ EndC(1),

where φTL: A⊗n → 1 is defined as in (2.1).

10 ALEXIS VIRELIZIER

Definition 2.1. By aKirby element of C, we shall mean a morphism α ∈ HomC(1, A)such that, for any framed link L, τC(L;α) is well-defined and invariant under iso-topies and 2-handle slides of L. A Kirby element α of C is said to be normalizedif τC(©

±1;α) is invertible in EndC(1), where ©±1 denotes the unknot with fram-ing ±1.

Note that the unit ηA : 1→ A of the categorical Hopf algebra A is a normalizedKirby element. The invariant of framed links associated ηA is the trivial one, thatis, τC(L; ηA) = 1 for any framed link L.

Lemma 2.2. Let α be a Kirby element of C. Then τC(L⊔L′;α) = τC(L;α) τC(L

′;α)for any framed link L and L′, where L⊔L′ denotes the disjoint union of L and L′.

Proof. Let TL (resp. TL′) be a special tangle such that L (resp. L′) is isotopicTL ◦ (∪− ⊗ · · · ⊗ ∪−) (resp. TL′ ◦ (∪− ⊗ · · · ⊗ ∪−)). The tangle T = TL ⊗ TL′ isthen special and such that the disjoint union L⊔L′ is isotopic T ◦ (∪− ⊗ · · · ⊗∪−).Therefore φT = φTL

⊗ φTL′ and so τC(L ⊔ L′;α) = τC(L;α) τC(L′;α). �

Let Θ± : A→ 1 be the morphism defined by

(2.2) Θ± ◦ iX = evX(idX∗ ⊗ θ±1X ),

where X is any object of C (see Figure 5). Note that if α be a Kirby element of C,then τC(©

±1;α) = Θ±α.

PSfrag replacements

iXiX

XX XXXX XX

Θ−Θ+

== AA

Figure 5.

Recall (see [Lic97]) that every closed, connected, and oriented 3-manifold canbe obtained from S3 by surgery along a framed link L ⊂ S3. In this paper, allconsidered 3-manifolds are supposed to be closed, connected, and oriented. For anyframed link L in S3, we will denote by ML the 3-manifold obtained from S3 bysurgery along L, by nL the number of components of L, and by b−(L) the numberof negative eigenvalues of the linking matrix of L.

Normalized Kirby elements are of special interest due to the following proposi-tion.

Proposition 2.3. Let α be a normalized Kirby element of C. Then

τC(ML;α) = (Θ+α)b−(L)−nL (Θ−α)

−b−(L) τC(L;α)

is an invariant of 3-manifolds. Moreover τC(M#M ′;α) = τC(M ;α) τC(M′;α) for

any 3-manifold M and M ′.

Proof. The fact that τC(ML;α) is an invariant of 3-manifolds follows from Kirbytheorem [Kir78]. Indeed τC(L;α), b−(L) and nL are invariant under 2-handleslides and τC(©

±1 ⊔ L;α) = (Θ±α) τC(L;α) by Lemma 2.2, b−(©1 ⊔ L) = b−(L),

b−(©−1 ⊔ L) = b−(L) + 1, and n©±1⊔L = nL + 1.


Let L and L′ be framed links in S3. The disjoint union L ⊔ L′ is then a framedlink in S3 such that ML⊔L′ ≃ ML#ML′ . Then the multiplicativity of τC(M ;α)with respect to the connected sum of 3-manifolds follows from Lemma 2.2 and theequalities b−(L ⊔ L′) = b−(L) + b−(L

′) and nL⊔L′ = nL + nL′ . �

Remarks 2.4. [a]. For any normalized Kirby element α of C, we have that

τC(S3;α) = 1 and τC(S

1 × S2;α) = (Θ+α)−1 εAα.

[b]. The invariant of 3-manifolds associated to the unit ηA : 1 → A of thecategorical Hopf algebra A (which is a normalized Kirby element) is the trivial one,that is, τC(M ; ηA) = 1 for any 3-manifold M .

In general, determining when a morphism A⊗n → 1 is of the form φT for somespecial n-tangle T is a quite difficult problem. Hence does so the problem of de-termining all the (normalized) Kirby elements of C. In the next section, we char-acterize a class of (normalized) Kirby elements of C by means of the structuralmorphisms of the categorical Hopf algebra A. This class will be shown to be suffi-ciently large to contain the Lyubashenko invariant (which is a categorical version ofthe Hennings-Kauffman-Radford invariant) and the Reshetikhin-Turaev invariant(computed from a semisimple quotient of C) when these are well-defined.

2.3. A class of Kirby elements. Let C be a ribbon k-category such that thefunctor (1.16) admits a coend (A, i).

Recall the notion of negligible morphisms (see Section 1.4). Set:

I(C) = {α ∈ HomC(1, A) | SAα− α ∈ NeglC(1, A) and

Γl(α ⊗ α)− α⊗ α ∈ NeglC(1, A⊗A)},

I(C)norm = {α ∈ I(C) | Θ+α 6= 0 and Θ−α 6= 0},

where Γl : A⊗A→ A⊗A and Θ± : A→ 1 are defined in (1.25) and (2.2).Note that, in the above definitions, the morphism Γl can be replaced by the

morphism Γr which is defined in (1.26) (see Corollary 1.2).Remark that the sets I(C) and I(C)norm always contain a non-zero element,

namely the unit ηA : 1→ A. Moreover,

(2.3) kI(C) + NeglC(1, A) ⊂ I(C) and k∗I(C)norm +NeglC(1, A) ⊂ I(C)norm,

since NeglC is a two-sided ⊗-ideal of C and NeglC(1, 1) = 0 (because EndC(1) = kis a field).

Theorem 2.5. (a) The elements of I(C) are Kirby elements of C.(b) The elements of I(C)norm are normalized Kirby elements of C.

Remarks 2.6. [a]. It follows from Proposition 2.3 that any α ∈ I(C)norm leadsto a 3-manifolds invariant τC(M ;α) with values in EndC(1) = k. Recall that thisinvariant is multiplicative with respect to the connected sum of 3-manifolds andverifies

τC(S3;α) = 1, τC(S

1 × S2;α) = (Θ+α)−1 εAα, τC(M ; ηA) = 1,

for any 3-manifold M .

[b]. Let α ∈ I(C)norm, n ∈ NeglC(1, A), and k ∈ k∗. Then

kα+ n ∈ I(C)norm and τC(M ; kα+ n) = τC(M ;α)


for any 3-manifold M . This follows from (2.3) and the choice of the normalizationin the definition of τC(M ;α) (see Proposition 2.3).

[c]. Suppose that C is 3-modular (see [Lyu95a]). Then A admits a (non-zero)two-sided integral λ ∈ HomC(1, A), that is,

mA(λ⊗ idA) = ηAεA = mA(idA ⊗ λ).

Moreover λ ∈ I(C)norm and τC(M ;λ) is the Lyubashenko invariant of 3-manifolds.Note that when C admits split idempotents, then λ is unique (up to scalar multiple),see [BKLT00].

The condition that A posses a (non-zero) two-sided integral is quite limitative(for example, when C is the category repH of representations of a finite-dimensionalHopf algebra H , this implies that H must be unimodular). In Section 5, we give anexample of non-unimodular ribbon Hopf algebra H and of an element α ∈ I(repH)which is not a two-sided integral and leads to a non-trivial invariant.

[d]. When C is finitely semisimple, we show in Section 3.3 (see Corollary 3.7) thatthere exists αC ∈ I(C) corresponding to the Reshetikhin-Turaev invariant definedwith C. Note that αC ∈ I(C) is not in general a two-sided integral.

More generally, we show in Section 3.5 (see Corollary 3.9) that I(C) containselements corresponding to the Reshetikhin-Turaev invariants defined with finitelysemisimple full ribbon subcategories of the ribbon semisimple quotient of C.

Proof of Theorem 2.5. Let us prove Part (a). Fix α ∈ I(C). Let L = L1 ∪ · · · ∪ Ln

be a framed link. Firstly, since SAα − α ∈ NeglC(1, A), then τC(L;α) does notdepend on choice of TL neither the orientation of L and is an isotopic invariantof the framed link L. Indeed, this is proved in the case SAα = α in [Lyu95a,Proposition 5.2.1]. The same arguments work when SAα − α ∈ NeglC(1, A) sinceNeglC is a two-sided ⊗-ideal of C.

Let us show that τC(L;α) is invariant under 2-handle slides. Choose an ori-entation for L. Without loss of generality, we can suppose that the componentL1 slides over L2. Let L′

2 be a copy of L2 (following the framing) and set L′ =(L1#L

′2) ∪ L2 ∪ · · · ∪ Ln. We have to show that τC(L

′;α) = τC(L;α). Let TL bea n-special tangle such that L is isotopic TL ◦ (∪− ⊗ · · · ⊗ ∪−), where the ith cup(with clockwise orientation) corresponds to the component Li, see Figure 6(a). Let∆2(TL) be the (2n + 2, 0)-tangle obtained by copying the 2nd component of TL(following the framing) in such a way that the endpoints of the new component arebetween the 2nd and 3th bottom endpoints and between the 4th and 5th bottomendpoints of TL. A n-special tangle T ′ such that L′ is isotopic T ′ ◦ (∪− ⊗ · · · ⊗∪−)(where the ith cup corresponds to the ith component of L′) can be constructedfrom ∆2(TL) as in Figure 6(b). For example, if TL is the 3-special tangle depictedin Figure 6(c), then let T ′ be the 3-special tangle of Figure 6(d).By the equalities of Figure 2.3 where X1, . . . , Xn are any objects of C, and by theuniqueness of the factorisation through a coend, we get that

φT ′ = φTL◦ (Γl ⊗ idA⊗(n−2)).

Therefore, since Γl(α⊗ α)− α⊗ α ∈ NeglC(1, A⊗A), we get that

τC(L′;α) = φT ′α⊗n = φTL

(idA⊗2 ⊗ α⊗(n−2))Γl(α⊗ α) = φTLα⊗n = τC(L;α).

To show Part (b), it suffices to remark that τC(©±1;α) ∈ EndC(1) is invertible

since τC(©±1;α) = Θ±α 6= 0 and EndC(1) = k is a field. �


PSfrag replacements

TLL ∼

L1 L2 Ln

. . .

(a) L ∼ TL◦(∪−⊗· · ·⊗∪−)

PSfrag replacements . . .

∆2(TL)

(b) T ′

(c) TL (d) T ′

Figure 6.

2.4. Kirby elements via functors. Let us see that Kirby elements can be “pulledback” via ribbon functors.

Let A, B, and C be ribbon k-categories. Suppose that the functor (1.16) of Aadmits a coend (A, i) and that the functor (1.16) of B admits a coend (B, j). Letπ : A → C and ι : B → C be ribbon functors. Since A and B are categoricalHopf algebras and π and ι are ribbon functors, the objects π(A) and ι(B) are Hopfalgebras in C (with structure maps induced by π and ι respectively).

Proposition 2.7. Suppose that π is surjective, ι is full and faithful, and that thereexists a Hopf algebra morphism ϕ : ι(B) → π(A) such that π(iX) = ϕ ◦ ι(jY )for all objects X ∈ A and Y ∈ B with π(X) = ι(Y ). Let β ∈ HomB(1, B) andα ∈ HomA(1, A) such that π(α) = ϕ ◦ ι(β).

(a) If β ∈ I(B), then α ∈ I(A) and τA(L;α) = τB(L;β) for any framed link L.(b) If β ∈ I(B)norm, then α ∈ I(A)norm and τA(M ;α) = τB(M ;β) for any

3-manifold M .

Proof. Let us prove Part (a). Suppose that β ∈ I(B). Since the structure mapsof π(A) and ι(B) are induced by π and ι from those of A and B respectively, andsince ϕ : ι(B) → π(A) is a Hopf algebra morphism, we have:

π(SAα− α) = (Sπ(A) − idπ(A))π(α) = (Sπ(A) − idπ(A))ϕι(β)

= ϕ(Sι(B) − idι(B))ι(β) = ϕι(SBβ − β)


PSfrag replacements

T ′(X1,...,Xn)

TL(X1,X2⊗X1,X3,...,Xn)

X1X1 X1X1

X1X1X1X1

X1

X1

X2X2

X2X2X2X2

X3X3

X3X3X3X3

XnXn XnXn

XnXnXnXn

X2⊗X1

X2⊗X1

X2⊗X1

X2⊗X1

idX2⊗X1

idX2⊗X1

idX2⊗X1

idX2⊗X1 iX1

iX1

φTLφTL

iX2

iX3

iX3

iXn

iXniX2⊗X1 Γl

==

=

. . .. . .

. . . . . .

A

A

AAA

A

A

AAA

Figure 7.

and

π(ΓAl (α⊗ α)− α⊗ α) = (Γ

π(A)l − idπ(A)⊗2)(π(α) ⊗ π(α))

= (Γπ(A)l − idπ(A)⊗2)ϕι(β ⊗ β)

= ϕ(Γι(B)l − idι(B)⊗2)ι(β ⊗ β)

= ϕι(ΓBl (β ⊗ β)− β ⊗ β).

Now, since SBβ − β and ΓBl (β ⊗ β) − β ⊗ β are negligible in B, ι is full, and

trBq = trCq ◦ ι, we get that π(SAα − α) and π(ΓAl (α ⊗ α) − α ⊗ α) are negligible

in C. Hence, since π is surjective and trCq ◦ π = trAq , the morphisms SAα − α and

ΓAl (α⊗ α)− α⊗ α are negligible in A, that is, α ∈ I(A).Let L = L1 ∪ · · · ∪ Ln be a framed link in S3. Let TL be a n-special tangle

such that L is isotopic TL ◦ (∪− ⊗ · · · ⊗ ∪−), where the ith cup (with clockwiseorientation) corresponds to the component Li. Let Y1, . . . , Yn be any objects of B.Since π is surjective, there exists objets X1, . . . , Xn of A such that π(Xk) = ι(Yk).Recall that, by assumption, π(iXk

) = ϕι(jYk). Since ι is full and the domain and

codomain of the morphism π(φATL)ϕ⊗n of C are ι(B⊗n) and 1 = ι(1) respectively,


there exists a morphism ξ : B⊗n → 1 in B such that ι(ξ) = π(φATL)ϕ⊗n. Then

ι(φBTL

◦ (jY1 ⊗ · · · ⊗ jYn))= ι(TB

L(Y1,...,Xn))

= T CL(ι(Y1),...,ι(Yn))

= T CL(π(X1),...,π(Xn))

= π(TAL(X1,...,Xn)

)

= π(φATL)(π(iX1 )⊗ · · · ⊗ π(iXn

))

= π(φATL)(ϕι(jY1 )⊗ · · · ⊗ ϕι(jYn

))

= π(φATL)ϕ⊗n ◦ ι(jY1 ⊗ · · · ⊗ jYn

)

= ι(ξ ◦ (jY1 ⊗ · · · ⊗ jYn

)).

Therefore, since ι is faithful, φBTL◦ (jY1 ⊗ · · · ⊗ jYn

) = ξ ◦ (jY1 ⊗ · · · ⊗ jYn) and so,

by the uniqueness of the factorization through a coend, we get that φBTL= ξ, that

is, ι(φBTL) = π(φATL

)ϕ⊗n. Hence, since the maps EndA(1) = k → EndC(1) = k andEndB(1) = k→ EndC(1) = k induced by π and ι respectively are the identity of k,we have that

τC(L;α) = π(τC(L;α)) = π(φATLα⊗n)

= π(φATL)ϕ⊗nι(β⊗n)

= ι(φBTLβ⊗n) = ι(τB(L;β)) = τB(L;β).

Let us prove Part (b). Suppose that β ∈ I(B)norm. Let Y be any objet of B.Since π is surjective, there exists an object X of A such that π(X) = ι(Y ). Recallthat, by assumption, π(iX) = ϕ◦ι(jY ). Since ι is full and the domain and codomainof the morphism π(ΘA

±)ϕ of C are ι(B) and 1 = ι(1) respectively, there exists a

morphism ς± : B → 1 in B such that ι(ς±) = π(ΘA±)ϕ. We have that

ι(ς± ◦ ι(jY )) = π(ΘA±)ϕ ◦ ι(jY ) = π(ΘA

±iX) = π(evAX(idX∗ ⊗ θA±1X ))

= evCπ(X)(idπ(X)∗ ⊗ θC±1π(X)) = ι(evBY )(ι(idY ∗)⊗ ι(θBY )

±1)

= ι(evBY (idY ∗ ⊗ θB±1Y )) = ι(ΘB

± ◦ jY ).

Therefore, since ι is faithful, ΘB± ◦ jY = ς± ◦ ι(jY ) and so, by the uniqueness of

the factorization through a coend, we get that ΘB± = ς±, that is, ι(Θ

B±) = π(ΘA

±)ϕ.Then

ΘC±α = π(ΘA

±α) = π(ΘA±)ϕι(β) = ι(ΘB

±β) = ΘB±β.

Hence, since ΘB±β 6= 0 and by Part (a), we get that α ∈ I(A)norm and τA(M ;α) =

τB(M ;β) for any 3-manifold M . �

3. The case of semisimple ribbon categories

In this section, we focus on the case of semisimple categories C. We give necessaryconditions for being in I(C). In particular, we show that there exist (even in thenon-modular case) elements of I(C) corresponding to Reshetikhin-Turaev invariantscomputed with finitely semisimple ribbon subcategories of C. Moreover, we studyKirby elements coming from semisimplification of ribbon categories.


3.1. Semisimple categories. Recall that a category C admits (finite) direct sumsif, for any finite set of objects X1, . . . , Xn of C, there exists an object X andmorphisms pi : X → Xi such that, for any object Y and morphisms fi : Y → Xi,there is a unique morphism f : Y → X with pi ◦ f = fi for all i. The object X isthen unique up to isomorphism. We write X = ⊕iXi and f = ⊕ifi.

A k-category is abelian if it admits (finite) direct sums, every morphism hasa kernel and a cokernel, every monomorphism is the kernel of its cokernel, everyepimorphism is the cokernel of its kernel, and every morphism is expressible asthe composite of an epimorphism followed by a monomorphism. In particular, anabelian category admits a null object (which is unique up to isomorphism). Notethat a morphism of an abelian k-category which is both a monomorphism and anepimorphism is an isomorphism.

Let C be an abelian k-category. A non-null object U of C is simple if every non-zero monomorphism V → U is an isomorphism, and every non-zero epimorphismU → V is an isomorphism. Any non-zero morphism between simple objects is anisomorphism. An object U of C is scalar if EndC(V ) = k. Note that if k is alge-braically closed, then every simple object is scalar. An object of C is indecomposableif it cannot be written as a direct sum of two non-null objects. Note that everyscalar or simple object is indecomposable.

By a semisimple k-category, we shall mean an abelian k-category for which everyobject is a (finite) direct sum of simple objects. By a finitely semisimple k-cate-gory, we shall mean a semisimple k-category which has finitely many isomorphismclasses of simple objects. Note that in a semisimple k-category, every scalar orindecomposable object is simple.

In a semisimple ribbon k-category C, any negligible morphism is null. Therefore,for every object X,Y of C, the pairing

(3.1) HomC(X,Y )⊗HomC(Y,X) → k f ⊗ g 7→ trq(gf)

is non-degenerate. Note that this implies that the quantum dimension of a scalarobject of C is invertible.

Lemma 3.1. Let C be a finitely semisimple ribbon k-category whose simple objectsare scalar. Let Λ be a (finite) set of representatives of isomorphism classes of simpleobjects of C. Fix an object X of C. For any λ ∈ Λ, set

nλ = dimk HomC(λ,X) = dimk HomC(X,λ)

and let {fλi | 1 ≤ i ≤ nλ} be a basis of HomC(λ,X) and {gλi | 1 ≤ i ≤ nλ} be a basis

of HomC(X,λ) such that gλi fλj = δi,j idλ for all 1 ≤ i, j ≤ nλ (such basis exist since

the pairing (3.1) is non-degenerate). Then

idX =∑

λ∈Λ

∑

1≤i≤nλ

fλi g

λi .

Proof. Note that since C is semisimple with scalar simple objects, the HomC ’sk-spaces are finite-dimensional. Since the category C is semisimple, the composi-tion induces a k-linear isomorphism ⊕λ∈ΛHomC(X,λ) ⊗HomC(λ,X) → EndC(X).Therefore, for all λ ∈ Λ and 1 ≤ i, j ≤ nλ, there exist aλ,i,j ∈ k such that


idX =∑

λ∈Λ

∑1≤i,j≤nλ

aλ,i,j fλi g

λj . Let λ ∈ Λ and 1 ≤ i, j ≤ nλ. Then

δi,j idλ = gλi fλj = gλi idX fλ

j =∑

µ∈Λ

∑

1≤k,l≤nλ

aµ,k,l gλi f

µk g

µl f

λj

=∑

µ∈Λ

∑

1≤k,l≤nλ

aµ,k,l δλ,µ δi,k δj,l idµ = aλ,i,j idλ,

and so aλ,i,j = δi,j . Hence idX =∑

λ∈Λ

∑1≤i≤nλ

fλi g

λi . �

3.2. Kirby elements of finitely semisimple ribbon categories. Let B be afinitely semisimple ribbon k-category whose simple objects are scalar.

Note that under these assumptions, the k-space HomB(X,Y ) is finite-dimensionalfor any objects X,Y of B.

Denote by Λ a (finite) set of representatives of isomorphism classes of simpleobjects of B. We can suppose that 1 ∈ Λ. For any λ ∈ Λ, there exists a uniqueλ∨ ∈ Λ such that λ∗ ≃ λ∨. The map ∨ : Λ → Λ is an involution and 1∨ = 1. Recallthat dimq(λ) 6= 0 for any λ ∈ Λ.

Set

(3.2) B =⊕

λ∈Λ

λ∗ ⊗ λ ∈ B.

In particular, there exist morphisms pλ : B → λ∗ ⊗ λ and qλ : λ∗ ⊗ λ → B suchthat idB =

∑λ∈Λ qλpλ and pλqµ = δλ,µ idλ∗⊗λ. Let X be an object of B. Since B

is semisimple, we can write X = ⊕i∈Iλi, where I is a finite set and λi ∈ Λ. We set

(3.3) jX =∑

i∈I

qλi◦ (q∗i ⊗ pi) : X

∗ ⊗X → B,

where pi : X → λi and qi : λi → X are morphisms in B such that idX =∑

i∈I qipiand piqj = δi,j idλi

. Note that jX does not depend on the choice of such a familyof morphisms. Remark that jλ = qλ for any λ ∈ Λ. It is not difficult to verify that(B, j) is a coend of the functor (1.16) of B. By Section 1.6, the object B is a Hopfalgebra in B.

For any λ ∈ Λ, set

(3.4) eλ = jλcoevλ : 1→ B and fλ = evλpλ : B → 1.

Note that fλeµ = δλ,µdimq(λ) for any λ, µ ∈ Λ.

Lemma 3.2. (a) (eλ)λ∈Λ is a basis of the k-space HomB(1, B).(b) (fλ)λ∈Λ is a basis of the k-space HomB(B, 1).(c) For any λ, µ ∈ Λ,

SB(eλ) = eλ∨ , ηB = e1

, εB(eλ) = dimq(λ),

(idB ⊗ fµ)∆B(eλ) = δλ,µeλ, mB(eλ ⊗ eµ) =∑

ν∈Λ

Nνλ,µeν ,

where Nνλ,µ = dimk HomB(λ⊗ µ, ν).

Proof. Let us prove Part (a). For any λ ∈ Λ, the k-space HomB(1, λ∗ ⊗ λ) is

one-dimensional (since λ is scalar) with basis coevλ. Therefore there exists xλ ∈ ksuch that pλf = xλcoevλ, and so f = idBf =

∑λ∈Λ qλpλf =

∑λ∈Λ xλeλ. Hence

the family (eλ)λ∈Λ generates HomB(1, B). To show that it is free, suppose that∑λ∈Λ xλeλ = 0. Then, for any µ ∈ Λ, 0 =

∑λ∈Λ xλfµeλ = dimq(µ)xµ and so

xµ = 0 since dimq(µ) 6= 0.


PSfrag replacements

B

B

B

B

B

λ∗

λ∗

λ∗

λ

idλ∗

idλ∗

SB

eλ

eλ∨jλ∗

jλ∗

= = =

(a) SB(eλ) = eλ∨PSfrag replacements

B B

B

B

BB

B

µ

λλ

λ

pµ

= δλ,µ = δλ,µ eλ

eλ

∆B

fµ

jλ

jλ

qλ

=

(b) (idB ⊗ fµ)∆B(eλ) = δλ,µeλ

Figure 8.

Part (b) can be shown similarly. Let us prove Part (c). Let λ, µ ∈ Λ. Bydefinition, ηB = j

1

= j1

coev1

= e1

(since coev1

= id1

) and εB(eλ) = εBjλcoevλ =evλcoevλ = dimq(λ). The equalities SB(eλ) = eλ∨ and (idB ⊗ fµ)∆B(eλ) = δλ,µeλare shown in Figure 8(a) and 8(b) respectively. Write λ⊗µ = ⊕i∈Iλi. In particular,there exits morphisms pi : λ ⊗ µ → λi and qi : λi → λ ⊗ µ such that idλ⊗µ =∑

i∈I qipi and piqj = δi,j idλi. Recall that jλ⊗µ =

∑i∈I jλi

(q∗i ⊗pi). For any ν ∈ Λ,since HomB(λ⊗ µ, ν) ∼= ⊕i∈IHomB(λi, ν), we have that

Nνλ,µ = dimk HomB(λ⊗ µ, ν) =

∑

i∈I

δλi,ν .

Then the equality mB(eλ ⊗ eµ) =∑

ν∈ΛNνλ,µeν is shown in Figure 9. �

Since B is semisimple and so its negligible morphisms are null, we have that:

I(B) = {α ∈ HomB(1, A) | SAα = α and Γl(α⊗ α) = α⊗ α}.

Lemma 3.3. Let α =∑

λ∈Λ αλeλ ∈ HomB(1, B), where αλ ∈ k. Suppose thatα ∈ I(B). Set Λα = {λ ∈ Λ |αλ 6= 0}. Then α = α

1

∑λ∈Λα

dimq(λ)eλ. Moreover,

the set Λα is such that Λ∨α = Λα and mB(α ⊗ eλ) = dimq(λ)α = mB(eλ ⊗ α) for

all λ ∈ Λα.

Proof. By Lemma 3.2(c), since SB(α) = α, we have that αλ = αλ∨ for all λ ∈ Λand so Λ∨

α = Λα. Let µ, ν ∈ Λ. Since Γr(α⊗α) = α⊗α and by using Lemma 3.2(c),


PSfrag replacements

BB

BB

B B

B

BB

B

B

mB

eλ eµ

eνjλi

jλijλi

q∗i

jν

qi

qi

pipi

pi

jµ⊗λ

idµ⊗λ

idµ⊗λ

idλ⊗µidλ⊗µ

idλ⊗µ idλ⊗µ

jλ⊗µ

jλ⊗µ

µµ

µ µ

µ

ν

λ

λ

λ

λ λ

λi

λi

λiλi

λi λi

µ⊗ λ

µ⊗ λ

λ⊗ µλ⊗ µ

λ⊗ µ

λ⊗ µ

λ⊗ µ

=∑

i∈I

=∑

i∈I

=∑

i∈I

=∑

i∈I

=∑

ν∈Λ

(∑

i∈I

δλi,ν

)=

∑

ν∈Λ

Nνλ,µ

= =

Figure 9. ∆B(eλ ⊗ eµ) =∑

ν∈ΛNνλ,µeν

we have that

dimq(µ)dimq(ν)αναµ = (fν ⊗ fλ)(α⊗ α) = (fν ⊗ fµ)Γr(α⊗ α)

=∑

λ,ω∈Λ

αλαω (fν ⊗ fµ)Γr(eλ ⊗ eω)

=∑

λ,ω∈Λ

αλαω fνmB(eλ ⊗ (idB ⊗ fµ)∆B(eω))

=∑

λ∈Λ

αλαµ fνmB(eλ ⊗ eµ)

=∑

λ,ω∈Λ

αλαµNωλ,µ fνeω

=∑

λ∈Λ

αλαµNνλ,µdimq(ν),

and so

(3.5) dimq(µ)αµαν = αµ

∑

λ∈Λ

Nνλ,µαλ.

Note that N1

λ,µ = dimk HomB(λ ⊗ µ, 1) = dimk HomB(λ, µ∗) = δλ,µ∨ for all λ, µ ∈

Λ. Hence (3.5) gives that dimq(µ)αµα1 = αµαµ∨ = α2µ and so αµ = α

1

dimq(µ)


whenever αµ 6= 0, that is, α = α1

∑λ∈Λα

dimq(λ)eλ. Finally, for any µ ∈ Λα,

mB(α⊗ eµ) =∑

λ∈Λ

αλmB(eλ ⊗ eµ)

=∑

λ,ν∈Λ

Nνλ,µαλ eν

=∑

ν∈Λ

dimq(µ)ανeν by (3.5) since αµ 6= 0

= dimq(µ)α.

Likewise, using the fact that Γl(α ⊗ α) = α ⊗ α, on gets that mB(eµ ⊗ α) =dimq(µ)α. �

By Lemma 3.3, determining I(B) resumes to find the subsets E of Λ for which∑λ∈E dimq(λ)eλ belongs to I(B). In the next theorem, we show that among these

subsets, there are those corresponding to monoidal subcategories of B.

Theorem 3.4. Let D be a full ribbon and abelian subcategory of B. Let ΛD be a(finite) set of representatives of isomorphism classes of simple objects of D. Wecan suppose that ΛD ⊂ Λ. Then

∑λ∈ΛD

dimq(λ)eλ ∈ I(B).

Remarks 3.5. [a]. We do not know if every element of I(B) is of this form.

[b]. In Section 3.3, we verify that∑

λ∈ΛDdimq(λ)eλ leads to the Reshetikhin-

Turaev invariant defined with D.

[c]. When B is modular in the sense that the pairing ωB : B⊗B → 1 defined in(1.24) is non-degenerate or, equivalently, that the S-matrix [trq(cµ,λ ◦ cλ,µ)]λ,µ∈Λ

is inversible, then∑

λ∈Λ dimq(λ)eλ is a two-sided integral of B (see [Ker97]) andso belongs to I(B). Nevertheless,

∑λ∈Λ dimq(λ)eλ is not in general a two-sided

integral of B.

Proof of Theorem 3.4. Firstly, since Λ∨D = ΛD and by Lemma 3.2(c),

SB(αD) =∑

λ∈ΛD

dimq(λ)SB(eλ) =∑

λ∈ΛD

dimq(λ∨)eλ∨ = αD.

Secondly, to show that Γr(αD ⊗ αD) = αD ⊗ αD, it suffices to show that, for anyλ ∈ ΛD,

(3.6) Γr(αD ⊗ eλ) = αD ⊗ eλ.

Fix λ ∈ ΛD. Let µ, ν ∈ ΛD and set nµ,ν = dimk HomB(ν ⊗ λ, µ). Since D is a fullsubcategory of B and the pairing

g ⊗ f ∈ HomB(ν ⊗ λ, µ)⊗HomB(µ, ν ⊗ λ) 7→ trq(gf) ∈ k

is non-degenerate, there exist basis {fµ,νi | 1 ≤ i ≤ nµ,ν} of HomB(µ, ν ⊗ λ) and

{gµ,νi | 1 ≤ i ≤ nµ,ν} of HomB(ν ⊗ λ, µ) such that gµ,νj fµ,νi = δi,j idµ for all 1 ≤

i, j ≤ nµ,ν . By Lemma 3.1, we have that

(3.7)∑

µ∈Λ

∑

1≤i≤nµ,ν

fµ,νi gµ,νi = idν⊗λ.


Let µ, ν ∈ ΛD. For any 1 ≤ i ≤ nµ,ν , set

F ν,µi = (gµ,νi ⊗ idλ)(idν ⊗ coevλ) : ν → µ⊗ λ∗,

Gν,µi = (idν ⊗ evλ)(f

µ,νi ⊗ idλ) : µ⊗ λ∗ → ν.

Since (λ∗, evλ, coevλ) is a right duality for λ, we have that {F ν,µi | 1 ≤ i ≤ nµ,ν} is a

basis for HomB(ν, µ⊗ λ∗) and {Gν,µi | 1 ≤ i ≤ nµ,ν} is a basis for HomB(µ⊗ λ∗, ν).

For any 1 ≤ i, j ≤ nµ,ν , since Gν,µj F ν,µ

i ∈ EndB(ν) and ν is scalar, we have that

dimq(ν)Gν,µj F ν,µ

i = trq(Gν,µj F ν,µ

i ) idν = trq(Fν,µi Gν,µ

j ) idν

= trq((gµ,νi ⊗ idλ)(idν ⊗ coevλ)(idν ⊗ evλ)(f

µ,νj ⊗ idλ)) idν

= trq(gµ,νi fµ,ν

j ) idν = trq(δi,j idµ) idν

= δi,j dimq(µ) idν .

Therefore, by Lemma 3.1,

(3.8)∑

ν∈Λ

∑

1≤i≤nµ,ν

dimq(ν)Fν,µi Gν,µ

i = dimq(µ) idµ⊗λ∗ .

Finally one gets (3.6) by the equalities depicted in Figure 10 which follow from thedinaturality of j, the definition of mB and ∆B , and equalities (3.7) and (3.8). �

Recall that an object X of B is invertible if X∗ ⊗X is isomorphic to 1.

Corrolary 3.6. Suppose that every simple object of B is invertible or, equivalently,that (Λ,⊗, 1) is a group. Then I(B) is composed by the scalar multiples of αG =∑

λ∈G dimq(λ)eλ, where G is any subgroup of Λ.

Proof. Let G be a subgroup of Λ. The full abelian subcategory D of B generatedby G is ribbon. Therefore, by Theorem 3.4, the element αG =

∑λ∈G dimq(λ)eλ

(and so its scalar multiples) belongs to I(B).Conversely, let α =

∑λ∈Λ αλeλ ∈ I(B). If α = 0, then α = 0 · αΛ. Suppose that

α 6= 0. By Lemma 3.3, α = α1

∑λ∈Λα

dimq(λ)eλ where Λα = {λ ∈ Λ |αλ 6= 0}.Then 1 ∈ Λα. Since every element of Λ is invertible, we have that Nν

λ,µ = δν,λ⊗µ

for any λ, µ, ν ∈ Λ. Let µ, ν ∈ Λα. By using (3.5), 0 6= dimq(µ)αµαν = αµαν⊗µ∨

and so αν⊗µ∨ 6= 0, that is, ν ⊗ µ∨ ∈ Λα. Hence Λα is a subgroup of Λ. �

3.3. On the Reshetikhin-Turaev invariant. Let B be a finitely semisimple rib-bon k-category whose simple objects are scalar. Let Λ be a (finite) set of represen-tatives of isomorphism classes of simple objects containing 1.

Set ∆± =∑

λ∈Λ v±1λ dimq(λ)

2 ∈ k, where vλ ∈ k is the (invertible) scalar definedby θλ = vλ idλ. Recall (see [Tur94, Bru00a]) that the Reshetikhin-Turaev invariantof 3-manifolds is well-defined when ∆+ 6= 0 6= ∆−. Moreover, if L is a framed linkin S3, it is given by:

RTB(ML) = ∆b−(L)−nL

+ ∆−b−(L)−

∑

c∈Col(L)

( n∏

j=1

dimq(c(Lj)))F (L, c) ∈ k.

Here Col(L) is the set of functions c : {L1, . . . , Ln} → Λ, where L1, . . . , Ln are thecomponents of L, and F (L, c) ∈ EndB(1) = k is the morphism represented by aplane diagram of L where the component Lj is colored with the object c(Lj).

By Theorem 3.4, αB =∑

λ∈Λ dimq(λ)eλ ∈ I(B), where eλ is defined as inSection 3.2.


PSfrag replacements

BB

BB

BB

BB

BB

BBBB

BBBB

ν

ν

ν

νν

λλ

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ

λ⊗ ν

λ⊗ ν

ν ⊗ λ

ν ⊗ λ

ν ⊗ λ

ν ⊗ λ

µ

µµ

µµ

µ

idλ⊗ν

idλ⊗ν

idν⊗λ

idν⊗λ idν⊗λ

SB

jλ⊗ν

jν⊗λ jν⊗λ

jµ

jµ

jµ

jλ

jλ

jλjλ

jλjλ

eλ

eλ

Γr

αD

αD

fµ,νi

fµ,νi

gµ,νi

gµ,νi

F ν,µi

Gν,µi

=∑

µ∈ΛD

dimq(µ)

Γr(αD ⊗ eλ) =∑

ν∈Λ

dimq(ν)

=∑

ν∈ΛD

dimq(ν)

=∑

ν∈ΛD

dimq(ν)

=∑

µ,ν∈ΛD

1≤i≤nµ,ν

dimq(ν)

=∑

µ,ν∈ΛD

1≤i≤nµ,ν

dimq(ν)

=∑

µ∈ΛD

∑

ν∈ΛD

1≤i≤nµ,ν

dimq(ν)

=

Figure 10. Γr(αD ⊗ eλ) = αD ⊗ eλ

Corrolary 3.7. Suppose that ∆± 6= 0. Then αB ∈ I(B)norm and τB(M ;αB) =RTB(M) for any 3-manifold M .

Proof. We have that

Θ±αB =∑

λ∈Λ

dimq(λ) evλ(idλ∗ ⊗ θ±1λ )coevλ =

∑

λ∈Λ

v±1λ dimq(λ)

2 = ∆± 6= 0.

Therefore, since αB ∈ I(B), one gets that αB ∈ I(B)norm.Let L = L1 ∪ · · · ∪ Ln be a framed link in S3. Let TL be a n-special tangle

such that L is isotopic TL ◦ (∪− ⊗ · · · ⊗ ∪−), where the ith cup (with clockwise


orientation) corresponds to the component Li. Then, for any c ∈ Col(L),

F (L, c) = φTL◦(jc(L1)coevc(L1)⊗· · ·⊗jc(Ln)coevc(Ln)) = φTL

◦(ec(L1)⊗· · ·⊗ec(Ln)).

Therefore

∑

c∈Col(L)

( n∏

j=1

dimq(c(Lj)))F (L, λ)

=∑

c∈Col(L)

φTL◦(dimq(c(L1)) ec(L1) ⊗ · · · ⊗ dimq(c(Ln))ec(Ln)

)

= φTL◦(∑

λ1∈Λ

dimq(λ1)eλ1 ⊗ · · · ⊗∑

λn∈Λ

dimq(λn)eλn

)

= φTL◦ (αB ⊗ · · · ⊗ αB).

Hence

RTB(ML) = ∆b−(L)−nL

+ ∆−b−(L)−

∑

λ∈Col(L)

( n∏

j=1

dimq(λ(Lj)))F (L, λ),

= (Θ+αB)b−(L)−nL (Θ−αB)

−b−(L) φTL◦ (αB ⊗ · · · ⊗ αB)

= τB(ML;αB).

�

3.4. Semisimplification of ribbon categories. Let C be a ribbon k-category.For any objects X,Y ∈ C, recall that NeglC(X,Y ) denotes the k-space of negligiblemorphisms of C fromX to Y (see Section 1.4). Let Cs be the category whose objectsare the same as in C, and whose morphisms are

(3.9) HomCs(X,Y ) = HomC(X,Y )/NeglC(X,Y )

for any objets X,Y ∈ Cs. The composition, monoidal structure, braiding, twist,and duality of Cs are induced by those of C.

When C has finite-dimensional Hom’s k-spaces, the category Cs is a semisim-ple ribbon k-category, called the semisimplification of C, and the simple objectsof Cs are the indecomposable objects of C with non-zero quantum dimension,see [Bru00b].

Let π : C → Cs be the functor defined by π(X) = X and π(f) = f +NeglC(X,Y )for any object X and any morphism f : X → Y in C. This is a surjective ribbonfunctor. Note that π is bijective on the objects.

3.5. Kirby elements from semisimplification. Let C be a ribbon k-categorywhich admits a coend (A, i) for the functor (1.16) and whose Hom’s spaces arefinite-dimensional. Denote by Cs the semisimplification of C and let π : C → Cs beits associated surjective ribbon functor (see Section 3.4).

Let B be a full ribbon and abelian subcategory of Cs which admits finitely manyisomorphism classes of simple objects, and whose simple objects are scalar. LetΛ be a (finite) set of representatives of isomorphism classes of simple objects of Bcontaining 1. For any object X of Cs, we denote by π−1(X) the (unique) object ofC such that π(π−1(X)) = X .

Let B = ⊕λ∈Λλ∗ ⊗ λ. In particular, there exist morphisms pλ : B → λ∗ ⊗ λ and

qλ : λ∗ ⊗ λ→ B of B such that idB =∑

λ∈Λ qλpλ and pλqµ = δλ,µ idλ∗⊗λ. For any


object X of B, we let jX : X∗ ⊗X → B as in (3.3). Recall that (B, j) is a coendfor the functor (1.16) of B and that jλ = qλ for any λ ∈ Λ (see Section 3.2).

Since B is a full ribbon subcategory of Cs and π is a ribbon functor, the objectsB and π(A) are Hopf algebra in Cs. Set

(3.10) ϕ =∑

λ∈Λ

π(iπ−1(λ))pλ ∈ HomCs(B, π(A)).

Lemma 3.8. ϕ : B → π(A) is a Hopf algebra morphism such that π(iX) = ϕjπ(X)

for all object X of C with π(X) ∈ B.

Proof. Let X be an object of C such that π(X) ∈ B. Since B is semisimple, wecan write π(X) = ⊕k∈Kλk, where K is a finite set and λk ∈ Λ. Recall thatjπ(X) =

∑k∈K qλk

(q∗k ⊗ pk), where pk : π(X) → λk and qk : λk → π(X) aremorphisms in B such that idπ(X) =

∑k∈K qkpk and pkql = δk,l idλk

. For any k ∈ K,

since π surjective, there exist morphisms Pk : X → π−1(λk) and Qk : π−1(λk) → Xin C such that π(Pk) = pk and π(Qk) = qk. Then, using the dinaturality of i andsince the functor π is ribbon,

ϕjπ(X) =∑

λ∈Λ

∑

k∈K

π(iπ−1(λ))pλqλk(q∗k ⊗ pk)

=∑

λ∈Λ

∑

k∈Kλk=λ

π(iπ−1(λ))(q∗k ⊗ pk)

=∑

λ∈Λ

∑

k∈Kλk=λ

π(iπ−1(λ)(Q

∗k ⊗ Pk)

)

=∑

λ∈Λ

∑

k∈Kλk=λ

π(iX(idX∗ ⊗QkPk)

)

= π(iX)(idπ(X)∗ ⊗∑

k∈K

qkpk)

= π(iX)(idπ(X)∗ ⊗ idπ(X)) = π(iX).

Let us verify that ϕ is a Hopf algebra morphism. For any objects X,Y of aribbon category, set

βX = (evX ⊗ idX∗∗)(idX ⊗ coevX∗) : X≃−→ X∗∗

and

γX,Y =(evX∗ ⊗ id(Y ⊗X)∗)(idX∗ ⊗ evY ∗ ⊗ id(Y⊗X)∗)

(idX∗⊗Y ∗ ⊗ coevY⊗X) : X∗ ⊗ Y ∗ ≃−→ (Y ⊗X)∗.

Let λ, µ ∈ Λ. Set U = π−1(λ) and V = π−1(µ). Then

επ(A)ϕjλ = π(εAiU ) = π(evU ) = evλ = εBjλ,

∆π(A)ϕjλ = π(∆AiU )

= π((iU ⊗ iU )(idU∗ ⊗ coevU ⊗ idU ))

= (ϕjλ ⊗ ϕjλ)(idλ∗ ⊗ coevλ ⊗ idλ) = (ϕ⊗ ϕ)∆Bjλ,


mπ(A)(ϕjλ ⊗ ϕjµ) = π(mA(iU ⊗ iV ))

= π(iV ⊗U (γU,V ⊗ idV ⊗U )(idU∗ ⊗ cU,V ∗⊗V ))

= ϕjµ⊗λ(γλ,µ ⊗ idµ⊗λ)(idλ∗ ⊗ cλ,µ∗⊗µ))

= ϕmB(jλ ⊗ jµ),

and

Sπ(A)ϕjλ = π(SAiU ) = π(iU∗(βU ⊗ θ−1U∗ )cU∗,U )

= ϕjλ∗(βλ ⊗ θ−1λ∗ )cλ∗,λ = ϕSBjλ.

Therefore, since idB =∑

λ∈Λ jλpλ, we get that επ(A)ϕ = εB, ∆π(A)ϕ = (ϕ⊗ϕ)∆B ,mπ(A)(ϕ⊗ϕ) = ϕmB, and Sπ(A)ϕ = ϕSB. Finally, we conclude by remarking thatϕηB = ϕj

1

= π(i1

) = π(ηA) = ηπ(A). �

Corrolary 3.9. Let β ∈ HomB(1, B) and α ∈ HomC(1, A) such that π(α) = ϕβ.

(a) If β ∈ I(B), then α ∈ I(C) and τC(L;α) = τB(L;β) for any framed link L.(b) If β ∈ I(B)norm, then α ∈ I(C)norm and τC(M ;α) = τB(M ;β) for any

3-manifold M .

Proof. This is an immediate consequence of Lemma 3.8 and Proposition 2.7. �

Remark 3.10. By Corollary 3.9, we have that⋃

B

π−1(ϕB(I(B))

)⊂ I(C) and

⋃

B

π−1(ϕB(I(B)

norm))⊂ I(C)norm,

where B runs over (equivalence classes of) finitely semisimple full ribbon and abeliansubcategories of Cs whose simple objects are scalar, and ϕB is the morphism (3.10)corresponding to B. We will see in Section 4 that these inclusions may be strict(see Remark 4.12). This means that the semisimplification process “lacks” someinvariants.

4. The case of categories of representations

In this section, we focus on the case of the category repH of representations ofa finite-dimensional ribbon Hopf algebra H . In particular, we describe I(repH) inpurely algebraic terms. One of the interest of such a description is to avoid therepresentation theory of H , which may be of wild type. Moreover, we show thatthe 3-manifolds invariants obtained with these Kirby elements can be computed byusing the Kauffman-Radford algorithm (even in the non-unimodular case).

4.1. Finite-dimensional Hopf algebras. All considered algebras are supposedto be over the field k.

Let H be a finite-dimensional Hopf algebra. Recall that a left (resp. right)integral for H is an element Λ ∈ H such that xΛ = ε(x)Λ (resp. Λx = ε(x)Λ) forall x ∈ H . A left (resp. right) integral for H∗ is then an element λ ∈ H∗ suchthat x(1)λ(x(2)) = λ(x) 1 (resp. λ(x(1))x(2) = λ(x) 1) for all x ∈ H . Since H isfinite-dimensional, the space of left (resp. right) integrals for H is one-dimensional,and there always exist non-zero right integral λ for H∗ and a non-zero left integralΛ for H such that λ(Λ) = λ(S(Λ)) = 1, see [Rad90, Proposition 3].

By [Rad94b, Corollary 2], the spaceH∗ endowed with the rightH-action defined,for any f ∈ H∗ and h, x ∈ H , by

(4.1) 〈f · h, x〉 = 〈f, hx〉,


is a free H-module of rank 1 with basis every non-zero right integral λ for H∗.Likewise, H endowed with the right H∗-action ↼ defined, for any f ∈ H∗ andx ∈ H , by

(4.2) x ↼ f = f(x(1))x(2),

is a free H∗-module of rank 1 with basis S(Λ), where Λ is a non-zero left integralfor H .

Lemma 4.1. Let λ be a right integral for H∗ and Λ be a left integral for H suchthat λ(Λ) = λ(S(Λ)) = 1. Let a ∈ H and f ∈ H∗. Then

(a) a = λ(aΛ(1))S(Λ(2)) = λ(S(Λ(2))a) Λ(1).(b) f = λ · a if and only if a = (f ⊗ S)∆(Λ).

Proof. Let us prove Part (a). Since λ is a right integral for H∗ and λ is a leftintegral for H such that λ(Λ) = 1, we have that

λ(aΛ(1))S(Λ(2)) = λ(a(1)Λ(1)) a(2)Λ(2) S(Λ(3))

= λ(a(1)Λ(1)) a(2)ε(Λ(2))

= λ(a(1)Λ) a(2)

= λ(Λ) ε(a(1))a(2)

= a.

The second equality of Part (a) can be proved similarly (by using the facts that λis a right integral for H∗ and S(Λ) is a right integral for H such that λ(S(Λ)) = 1).

Let us prove Part (b). If f = λ · a then, by Part (a),

a = λ(aΛ(1))S(Λ(2)) = f(Λ(1))S(Λ(2)) = (f ⊗ S)∆(Λ).

Conversely, if a = (f ⊗ S)∆(Λ) then, by Part (a),

λ(ax) = λ(f(Λ(1))S(Λ(2))x) = f(λ(S(Λ(2))x)Λ(1)) = f(x)

for all x ∈ H , and so f = λ · a. �

Recall that an element h ∈ H is grouplike if ∆(g) = g⊗g and ε(g) = 1. We denoteby G(H) the space of grouplike elements of H . Recall (see [Abe80]) that there exista unique grouplike element g of H such that x(1)λ(x(2)) = λ(x) g for any x ∈ H andany right integral λ ∈ H∗, and a unique grouplike element ν ∈ G(H∗) = Alg(H, k)such that Λx = ν(x)Λ for any x ∈ H and any left integral Λ ∈ H . The elementg ∈ H (resp. ν ∈ H∗) is called the distinguished grouplike element of H (resp.of H∗). The Hopf algebra H is said to be unimodular if its integrals are to sided,that is, if ν = ε.

By [Rad94b, Theorem 3] and [Rad94b, Proposition 3], right integrals for H∗ anddistinguished grouplike elements of H and H∗ are related by:

λ(xy) = λ(S2(y ↼ ν)x) = λ((S2(y)↼ ν)x) for all x, y ∈ H,(4.3)

λ(S(x)) = λ(gx) for all x ∈ H .(4.4)


4.2. Quasitriangular Hopf algebras. Following [Dri90], a Hopf algebra H isquasitriangular if it is endowed with an invertible element R ∈ H ⊗H (the R-ma-trix ) such that, for any x ∈ H ,

R∆(x) = σ∆(x)R,(4.5)

(idH ⊗∆)(R) = R13R12,(4.6)

(∆⊗ idH)(R) = R13R23.(4.7)

where σ : H ⊗H → H ⊗H denotes the usual flip map. The R-matrix verifies:

(ε⊗ idH)(R) = (idH ⊗ ε)(R) = 1,(4.8)

(S ⊗ idH)(R) = R−1 = (idH ⊗ S−1)(R),(4.9)

R23R13R12 = R12R13R23.(4.10)

The Drinfeld element u associated to R is u = m(S ⊗ idH)σ(R) ∈ H . It isinvertible and verifies:

u−1 = m(idH ⊗ S2)(R21),(4.11)

S2(x) = uxu−1 for all x ∈ H,(4.12)

∆(u) = (R21R)−1(u⊗ u),(4.13)

ε(u) = 1.(4.14)

Let H be a finite-dimensional quasitriangular Hopf algebra and let ν ∈ H∗ bethe distinguished grouplike element of H∗. Set hν = (idH ⊗ ν)(R) ∈ H . By (4.7)and (4.8), hν is grouplike.

4.3. Ribbon Hopf algebras. Following [RT91], a ribbon Hopf algebra is a qua-sitriangular Hopf algebra H endowed with an invertible element θ ∈ H (the twist)such that:

θ is central,(4.15)

S(θ) = θ,(4.16)

∆(θ) = (θ ⊗ θ)R21R.(4.17)

The twist verifies:

ε(θ1) = 1,(4.18)

θ−2 = uS(u) = S(u)u.(4.19)

Set G = θu ∈ H . Then G is grouplike and verifies:

S(u) = G−1uG−1,(4.20)

S2(x) = GxG−1.(4.21)

The element G is called the special grouplike element of H .By [Rad92, Theorem 2] and (4.20), the special grouplike element G of a finite-

dimensional ribbon Hopf algebraH is related to the distinguished grouplike elementg of H and to hν = (idH ⊗ ν)(R) ∈ G(H) by

(4.22) g = G2hν .

Relations (4.4) and (4.22) imply that

λ(S(x)) = λ(G2hνx) for all x ∈ H .(4.23)


4.4. Category of representations of a ribbon Hopf algebra. Let H be aribbon Hopf algebra with R-matrix R and twist θ. Denote by repH the k-categoryof finite-dimensional left H-modules and H-linear homomorphisms. The categoryrepH is a monoidal k-category with tensor product and unit object defined in theusual way using the comultiplication and counit of H . The category repH posses aleft duality: for any module M ∈ repH , set M∗ = Homk(M, k), where h ∈ H actsas the transpose of x ∈M 7→ S(h)x ∈M . The duality morphism evM :M∗⊗M →1 = k is the evaluation pairing and, if (ek)k is a basis of M with dual basis (e∗k)k,then coev(1k) =

∑k ek ⊗ e∗k. The category repH is braided: for modules M,N ∈

repH , the braiding cM,N :M ⊗N → N ⊗M is the composition of multiplication byR and the flip map σM,N :M ⊗N → N ⊗M . The category repH is ribbon: for anymodule M ∈ repH , the twist θM : M → M is the multiplication by θ. Recall, see(1.9) and (1.10), that repH posses a right duality M ∈ repH 7→ (M∗, evM , coevM ).Finally, the k-category repH is pure, that is, EndrepH

(k) = k. Hence repH is aribbon k-category in the sense of Section 1.2.

Lemma 4.2. Let G be the special grouplike element of H and M be a finite-dimensional left H-module. Then,

(a) evM (m⊗ f) = f(Gm) for any f ∈M∗ and m ∈M .(b) coevM = (idM∗ ⊗G−1idM )σM,M∗coevM .

Proof. Write R =∑

i ai ⊗ bi. Recall that u =∑

i S(bi)ai. Then

evM (m⊗ f) = evM cM,M∗(θM ⊗ idM∗)(m⊗ f)

=∑

ievM (bif ⊗ aiθm) =∑

if(S(bi)aiθm)

= f(uθm) = f(Gm).

Let (ek)k be a basis of M and (e∗k)k be its dual basis. Note that if g is any k-linearendomorphism of M , then

∑k g

∗(e∗k)⊗ ek =∑

k e∗k ⊗ g(ek). For any h ∈ H , denote

by ρ(h) the k-linear endomorphism of M defined by m ∈M 7→ hm ∈M . By (4.9)and (4.11), we have that R−1 =

∑i S(ai)⊗ bi and u

−1 =∑

i biS2(ai). Then

coevM (1k) = (idM∗ ⊗ θ−1M )(cM∗,M )−1coevM (1k)

=∑

k,iS(ai)e∗k ⊗ θ−1bi · ek

=∑

i

(idM∗ ⊗ ρ(θ−1bi)

)(∑kρ(S

2(ai))∗(e∗k)⊗ ek

)

=∑

i

(idM∗ ⊗ ρ(θ−1bi)

)(∑ke

∗k ⊗ ρ(S2(ai))(ek)

)

=(idM∗ ⊗ ρ(θ−1∑

ibiS2(ai))

)(∑ke

∗k ⊗ ek

)

=(idM∗ ⊗ ρ(θ−1u−1)

)σM,M∗coevM (1k)

=(idM∗ ⊗ ρ(G−1)

)σM,M∗coevM (1k)

= (idM∗ ⊗G−1idM )σM,M∗coevM (1k).

This completes the proof of the lemma. �

We immediately deduce from Lemma 4.2 that, in the category repH , we have

trq(f) = Tr(Gf) = Tr(G−1f),

dimq(M) = Tr(G idM ) = Tr(G−1idM ),

for all module M ∈ repH and all H-linear endomorphism f ofM , where Tr denotesthe usual trace of k-linear endomorphisms.


4.5. Braided Hopf algebra associated to ribbon Hopf algebras. Let H bea finite-dimensional ribbon Hopf algebra. The ribbon k-category repH of finite-dimensional left H-modules posses a coend (A, i) for the functor (1.16). Moreprecisely, A = H∗ = Homk(H, k) as a k-space and is endowed with the coadjointleft H-action ⊲ given, for any f ∈ A and h, x ∈ H , by

(4.24) 〈h ⊲ f, x〉 = 〈f, S(h(1))xh(2)〉,

where 〈, 〉 denotes the usual pairing between a k-space and its dual. Given a moduleM ∈ repH , the map iM :M∗ ⊗M → A is given by

(4.25) 〈iM (l ⊗m), x〉 = 〈l, xm〉,

for all l ∈M∗, m ∈M , and x ∈ H .

Lemma 4.3. If ξ is a dinatural transformation from the functor (1.16) of repH toa module Z ∈ repH , then the (unique) morphism r : A→ Z such that ξM = r ◦ iMfor all M ∈ repH is given by f ∈ A = H∗ 7→ r(f) = ξH(f ⊗ 1H).

Recall (see Section 1.6) that A is a Hopf algebra in repH . Using Lemma 4.3,the structural morphisms of A can be explicitly described in terms of the structuremaps of the Hopf algebra H . Nevertheless, it is more convenient to write downits pre-dual structural morphisms: for example, since A = H∗ as a k-space and His finite-dimensional, the pre-dual of the multiplication mA : A ⊗ A → A of A isa morphism ∆Bd : H → H ⊗ H such that (∆Bd)∗ = mA. That yields a k-spaceHBd = H endowed with a comultiplication ∆Bd : HBd → HBd ⊗ HBd, a counitεBd : HBd

1 → k, a unit ηBd : k→ HBd, a multiplication mBd : HBd ⊗HBd → HBd,and an antipode SBd : HBd → HBd. These structure maps, described in thefollowing lemma, verify the same axioms as those of a Hopf algebra except that theusual flip maps are replaced by the braiding of repH . The space HBd is called thebraided Hopf algebra associated to H , see [Maj93, Lyu95b].

Lemma 4.4 (cf. [Lyu95a]). The braided Hopf algebra HBd associated to H can bedescribed as follows :

• HBd = H as an algebra;• ∆Bd(x) =

∑i x(2)ai ⊗ S(bi(1))x(1)bi(2) =

∑i S(ai(1))x(1)ai(2) ⊗ S(bi)x(2),

• εBd = ε;• SBd

α (x) =∑

i S(ai)θ2S(x)ubi =

∑i S(ai)S(x)S(u)

−1bi,

for any x ∈ H, where R =∑

i ai ⊗ bi is the R-matrix, u the Drinfeld element, andθ the twist of H.

4.6. Kirby elements of ribbon Hopf algebras. Throughout this section, Hwill denote a finite-dimensional ribbon Hopf algebra with R-matrix R ∈ H ⊗ Hand twist θ ∈ H . Let u ∈ H be the Drinfeld element of H , G = θu be the specialgrouplike element of H , λ ∈ H∗ be a non-zero right integral for H∗, Λ ∈ H bea non-zero left integral for H such that λ(Λ) = λ(S(Λ)) = 1, g ∈ G(H) be thedistinguished grouplike element of H , ν ∈ G(H∗) = Alg(H, k) be the distinguishedgrouplike element of H∗, and hν = (idH ⊗ ν)(R) ∈ G(H). Let (A, i) be the coendfor the functor (1.16) of repH , and let HBd be the braided Hopf algebra associatedto H . Recall that A = H∗ endowed with the coadjoint left H-action ⊲ and thatA is a Hopf algebra in the category repH whose structure maps are dual to thoseof HBd. Recall that · denotes the right action of H on H∗ defined in (4.1) and ↼


denotes the right H∗-action on H defined in (4.2).

Set:

φ :

{H → Homk(k, H

∗)

z 7→ φz = (α 7→ αλ · z), T :

{H → H

z 7→ T (z) = (S(z)↼ ν)hν,

and

ψ :

{H⊗n → Homk(H

∗⊗n, k)

X 7→ ψX =(f1 ⊗ · · · ⊗ fn 7→ 〈f1 ⊗ · · · ⊗ fn, X〉

) .

Denote by ⊳ be the right action of H on H⊗n given by

(x1 ⊗ · · · ⊗ xn) ⊳ h = S(h(1))x1h(2) ⊗ · · · ⊗ S(h(2n−1))xnh(2n)

for any h ∈ H and x1, . . . , xn ∈ H . Set

L(H) = {z ∈ H | (x ↼ ν)z = zx for any x ∈ H},

N(H) = {z ∈ L(H) |λ(za) = 0 for any a ∈ Z(H)},

Vn(H) = {X ∈ H⊗n |X ⊳ h = ε(h)X for any h ∈ H}.

Note that if H is unimodular, then L(H) = Z(H) and T (z) = S(z) for all z ∈ H .By the definition of L(H) and by (4.3), we have that

(4.26) λ(zxy) = λ(zS2(y)x)

for all z ∈ L(H) and x, y ∈ H .

Lemma 4.5. (a) The map φ is an k-isomorphism with inverse given by:

α ∈ Homk(k, H∗) 7→ φ−1(α) = (α(1k)⊗ S)∆(Λ) ∈ H.

(b) The map φ induces a k-isomorphism between L(H) and HomrepH(k, A).

(c) The map φ induces a k-isomorphism between N(H) and NeglrepH(k, A).

(d) The map ψ induces a k-isomorphism between Vn(H) and HomrepH(A⊗n, k).

(e) V1(H) = Z(H).

Remark 4.6. In Appendix A, we use the space L(H) and the morphism T , de-fined from topological considerations, to parameterize all the traces of a finite-dimensional ribbon Hopf algebra H .

Proof. Let us prove Part (a). Since (H∗, ·) is a free right H-module of rank 1 withbasis λ, φ is an isomorphism. The expression of φ−1 follows from Lemma 4.1(b).

Let us prove Part (b). Let z ∈ H . We have to show that z ∈ L(H) if and onlyif φz is H-linear. Suppose that φz is H-linear. For all a, h ∈ H ,

ε(h)λ(za) = 〈ε(S−1(h))φz(1k), a〉 = 〈S−1(h) ⊲ φz(1k), a〉

= λ(zh(2)aS−1(h(1))) = λ(S2(S−1(h(1))↼ ν)zh(2)a) by (4.3),

and so, since (H∗, ·) is a free right H-module of rank 1 with basis λ,

(4.27) ε(h)z = S2(S−1(h(1))↼ ν)zh(2) = ν−1(h(2))S(h(1))zh(3)


for all h ∈ H . Hence, for any x ∈ H ,

(x ↼ ν)z = ν(x(1))x(2)z = ν(x(1))x(2)ε(x(3))z

= ν(x(1)) ν−1(x(4))x(2)S(x(3))zx(5) by (4.27)

= ν(x(1)) ν−1(x(3)) ε(x(2)) zx(4) = ν(x(1)) ν

−1(x(2)) zx(3)

= ε(x(1)) zx(2) = zx,

and so z ∈ L(H). Conversely, suppose that z ∈ L(H). Then, for any x, h ∈ H ,

λ(zS(h(1))xh(2)) = λ(zS2(h(2))S(h(1))x) by (4.26)

= λ(zS(h(1)S(h(2)))x) = ε(h)λ(zx),

and so φz is H-linear.Let us prove Part (d). Note that ψ is an isomorphism sinceH is finite-dimensional.

Let X ∈ H⊗n. For all h ∈ H and f1, . . . , fn ∈ H∗,

ψX

(h ⊲ (f1 ⊗ · · · ⊗ fn)

)= 〈f1 ⊗ · · · ⊗ fn, X ⊳ h〉

andε(h)ψX(f1 ⊗ · · · ⊗ fn) = 〈f1 ⊗ · · · ⊗ fn, ε(h)X〉.

Therefore ψX is H-linear if and only if X ∈ Vn(H).Let us prove Part (e). Let a ∈ Z(H). For all h ∈ H ,

a ⊳ h = S(h(1))ah(2) = S(h(1))h(2)a = ε(h) a

and so a ∈ V1(H). Conversely, let a ∈ V1(H). For all x ∈ H ,

xa = x(1)ε(x(2)) a = x(1)(a ⊳ x(2)) = x(1)S(x(2))ax(3) = ε(x(1)) ax(2) = ax,

and so a ∈ Z(H).Finally, let us prove Part (c). Let z ∈ L(H). Since EndrepH

(k) = k, we havethat φz is negligible if and only if ψaφz = λ(za) = 0 for all a ∈ V1(H) = Z(H),that is, if and only if z ∈ N(H). �

Lemma 4.7. (a) L(H) is a commutative algebra with product ∗ defined by

x ∗ z = λ(xS(z(2))) z(1) = λ(zS(x(2)))x(1)

= λ(z(1)S−1(x)) z(2) = λ(x(1)S

−1(z))x(2)

for any x, z ∈ L(H), and with S(Λ) as unit element.(b) For any z ∈ L(H), SA ◦ φz = φT (z), where SA denotes the antipode of the

categorical Hopf algebra A.(c) T induces on L(H) an involutory algebra-automorphism, that is,

T (x ∗ z) = T (x) ∗ T (z), T (S(Λ)) = S(Λ), and T 2(x) = x

for all x, z ∈ L(H).

Proof. Let us prove Part (a). Since A is a Hopf algebra in repH , the spaceHomrepH

(k, A) is an algebra for the convolution product α∗β = mA(α⊗β) and withunit element ηA. This algebra structure transports to L(H) via the k-isomorphismφ : L(H) → HomrepH

(k, A). Let x, z ∈ L(H). Then

x ∗ z = φ−1(φx ∗ φz) = φ−1(mA(φx ⊗ φz))

= 〈mA(φx ⊗ φz)(1k),Λ(1)〉S(Λ(2)) by Lemma 4.5(a)

= 〈λ · x⊗ λ · z,∆Bd(Λ(1))〉S(Λ(2))


Write R =∑

i ai ⊗ bi. By using Lemma 4.4, (4.8) and (4.26), we have that

x ∗ z =∑

iλ(xΛ(2)ai)λ(zS(bi(1))Λ(1)bi(2))S(Λ(3))

=∑

iλ(xΛ(2)ai)λ(zS2(bi(2))S(bi(1))Λ(1))S(Λ(3))

=∑

iλ(xΛ(2)aiε(bi))λ(zΛ(1))S(Λ(3)),

that is,

(4.28) x ∗ z = λ(xΛ(2))λ(zΛ(1))S(Λ(3)).

Likewise

x ∗ z =∑

iλ(xS(ai(1))Λ(1)ai(2))λ(zS(bi)Λ(2))S(Λ(3))

=∑

iλ(xS2(ai(2))S(ai(1))Λ(1))λ(zS(bi)Λ(2))S(Λ(3))

=∑

iλ(xΛ(1))λ(zS(ε(ai) bi)Λ(2))S(Λ(3)),

that is,

(4.29) x ∗ z = λ(xΛ(1))λ(zΛ(2))S(Λ(3)).

Now, by Lemma 4.1(a),

z = λ(zΛ(1))S(Λ(2)),(4.30)

x = λ(xΛ(1))S(Λ(2)),(4.31)

so that

z(1) ⊗ S−1(z(2)) = λ(zΛ(1))S(Λ(3))⊗ Λ(2)(4.32)

x(1) ⊗ S−1(x(2)) = λ(xΛ(1))S(Λ(3))⊗ Λ(2).(4.33)

Hence

x ∗ z = λ(xΛ(2))λ(zΛ(1))S(Λ(3)) by (4.28)

= λ(xS−1(z(2))) z(1) by (4.32)

= λ(xS(z(2))) z(1) by (4.26),

x ∗ z = λ(zΛ(1))λ(xΛ(2))S(Λ(3)) by (4.28)

= λ(zΛ(1))λ(x(1)Λ(2))x(2)Λ(3)S(Λ(4))

= λ(x(1)S−1(z))x(2) by (4.30),

x ∗ z = λ(xΛ(1))λ(zΛ(2))S(Λ(3)) by (4.29)

= λ(zS−1(x(2)))x(1) by (4.33)

= λ(zS(x(2)))x(1) by (4.26),

and

x ∗ z = λ(xΛ(1))λ(zΛ(2))S(Λ(3)) by (4.29)

= λ(xΛ(1))λ(z(1)Λ(2)) z(2)Λ(3)S(Λ(4))

= λ(z(1)S−1(x)) z(2) by (4.31).

Note that these expressions of the product of L(H) show that L(H) is commutative.Moreover, by Lemma 4.5(a), the unit element of L(H) is

φ−1(ηA) = (ηA(1)⊗ S)∆(Λ) = (ε⊗ S)∆(Λ) = S(Λ).


Let us prove Part (b). Let z ∈ L(H). Write R =∑

i ai ⊗ bi. For any x ∈ H ,

〈SAφz(1k), x〉 = 〈λ · z, SBd(x)〉

=∑

iλ(zS(ai)θ2S(x)ubi) by Lemma 4.4

=∑

iλ(zS2(u)S2(bi)S(ai)θ

2S(x)) by (4.26)

= λ(zu2θ2S(x)) since S2(u) = u and (S ⊗ S)(R) = R

= λ(zG2S(x)) = λ((G2 ↼ ν)zS(x)) = ν(G2)λ(G2zS(x))

= ν(G2)λ(G2hνxS−1(z)G−2) by (4.23)

= ν(G2)λ((G−2 ↼ ν)G2hνxS−1(x)) by (4.3)

= ν(G2) ν(G−2)λ(hνxS−1(z))

= λ((S(z)↼ ν)hνx) by (4.3)

= λ(T (z)x) = 〈φT (z)(1k), x〉,

that is SAφz = φT (z).Let us prove Part (c). Let z ∈ L(H). Firstly T (z) ∈ L(H) since φT (z) = SAφz

is H-linear. Moreover, since S2A = θA and the twist of repH is natural and verify

θk = idk, we have that

T 2(z) = φ−1(φT 2(z)) = φ−1(S2Aφz) = φ−1(θAφz) = φ−1(φzθk) = φ−1(φz) = z.

For any x, z ∈ L(H),

φT (x∗z) = SAφx∗z = SA(φx ∗ φz) = SAmA(φx ⊗ φz)

= mA(SA ⊗ SA)cA,A(φx ⊗ φz) = mA(SA ⊗ SA)(φz ⊗ φx)ck,k

= mA(SAφz ⊗ SAφx) = φT (z) ∗ φT (x) = φT (z)∗T (x)

and so T (x ∗ z) = T (z) ∗ T (x) = T (x) ∗ T (z). Finally T (S(Λ)) = S(Λ) sinceφT (S(Λ)) = SAφS(Λ) = SAηA = ηA = φS(Λ). �

In the next theorem, we describe the sets I(repH) and I(repH)norm in algebraicterms. Set

(4.34) I(H) = φ−1(I(repH)) and I(H)norm = φ−1(I(repH)norm),

where φ : L(H) → HomrepH(k,A) is as in Lemma 4.5. Note that I(H)norm ⊂ I(H).

It follows from (2.3) that

kI(H) +N(H) ⊂ I(H) and k∗I(H)norm +N(H) ⊂ I(H)norm.

Since ηA ∈ I(repH)norm and by Lemma 4.7, the right integral S(Λ) for H belongsto I(H) and I(H)norm.

Theorem 4.8. I(H) is constituted by the element z ∈ L(H) verifying

(a) T (z)− z ∈ N(H), that is, λ(T (z)a) = λ(za) for all a ∈ Z(H);(b)

∑iλ(zxi(1))λ(zxi(2)yi) =

∑iλ(zxi)λ(zyi) for all X =

∑i xi ⊗ yi ∈ V2(H).

In particular, I(H) contains the elements z ∈ L(H) satisfying T (z) = z andλ(zx(1))zx(2) = λ(zx)z for all x ∈ H. Moreover, an element z ∈ I(H) belongs

to I(H)norm if and only if λ(zθ) 6= 0 6= λ(zθ−1).

Note that the sets I(H) and I(H)norm do not depend on the choice of the non-zero right integral λ for H∗. In Section 5, we give an example of determination ofthese sets for a family of non-unimodular ribbon Hopf algebras.


Proof. Let z ∈ L(H). By Lemma 4.7(b), we have that SAφz = φT (z). Therefore,using Lemma 4.5(c), we get that SAφz−φz ∈ NeglrepH

(k, A) if and only if T (z)−z ∈

N(H) , that is, if and only if λ(T (z)a) = λ(za) for all a ∈ Z(H). Note that thislast property is in particular satisfied when T (z) = z.

Since EndrepH(k) = k, the morphism Γr(φz ⊗ φz) − φz ⊗ φz : k → A ⊗ A is

negligible if and only ifG◦(Γr(φz⊗φz)−φz⊗φz

)= 0 for anyG ∈ HomrepH

(A⊗A, k).By Lemma 4.5(d), this is equivalent to ψX(Γr(φz ⊗ φz) − φz ⊗ φz) = 0 for allX ∈ V2(H). Now, writing R =

∑i ai ⊗ bi and using Lemma 4.4, we have that for

any x, y ∈ H ,

〈Γr(φz ⊗ φz)(1k), x⊗ y〉

= 〈(mA ⊗ idA)(idA ⊗∆A)(λ · z ⊗ λ · z), x⊗ y〉

= 〈(idA ⊗∆A)(λ · z ⊗ λ · z),∆Bd(x) ⊗ y〉

=∑

i〈(idA ⊗∆A)(λ · z ⊗ λ · z), S(ai(1))x(1)ai(2) ⊗ S(bi)x(2) ⊗ y〉

=∑

i〈λ · z ⊗ λ · z, S(ai(1))x(1)ai(2) ⊗ S(bi)x(2)y〉

=∑

iλ(zS(ai(1))x(1)ai(2))λ(zS(bi)x(2)y)

=∑

iλ(zS2(ai(2))S(ai(1))x(1))λ(zS(bi)x(2)y) by (4.26)

=∑

iλ(zx(1))λ(zS(ε(ai)bi)x(2)y)

= λ(zx(1))λ(zx(2)y) by (4.8).

Therefore the morphism Γr(φz ⊗φz)−φz ⊗φz : k→ A⊗A is negligible if and onlyif λ(zxi(1))λ(zxi(2)yi) = λ(zxi)λ(zyi) for all X =

∑xi ⊗ yi ∈ V2(H). Note that

this last property is in particular satisfied when λ(zx(1))λ(zx(2)y) = λ(zx)λ(zy)for all x, y ∈ H , that is, since (H∗, ·) is a free right H-module of rank 1 with basisλ, when λ(zx(1)) zx(2) = λ(zx) z for all x ∈ H .

Finally, by using Lemma 4.3, we have that:

Θ±φz = evH(idH∗ ⊗ θ±1H )(φz(1k)⊗ 1H) = evH(λ · z ⊗ θ±1) = λ(zθ±1).

This completes the proof of the theorem. �

Corrolary 4.9. 1 ∈ I(H) if and only if H is unimodular.

Proof. Suppose that H is unimodular. Therefore L(H) = Z(H) and T (z) = S(z)for all z ∈ Z(H). In particular 1 ∈ L(H) and T (1) = 1. Moreover, λ(x(1))x(2) =λ(x)1 for all x ∈ H (since λ is a right integral for H∗). Hence 1 ∈ I(H) byTheorem 4.8.

Conversely, suppose that 1 ∈ I(H). In particular 1 ∈ L(H) and so z ↼ ν = zfor all z ∈ H . Therefore ε(z) = ε(z ↼ ν) = ν(z(1))ε(z(2)) = ν(z) for all z ∈ H , thatis, H is unimodular. �

4.7. Elements elements from semisimplification. Let H be a finite-dimen-sional ribbon Hopf algebra. Let (A, i) be the coend of the functor (1.16) of repH(as in Section 4.5). Denote by repsH the semisimplification of repH and by π itsassociated surjective ribbon functor repH → repsH (see Section 3.4). Let φ : L(H) →HomrepH

(k, A) be as is Section 4.6. Set

(4.35) I(H)s =⋃

B

φ−1(π−1

(ϕB(I(B))

)),


where B runs over (equivalence classes of) finitely semisimple full ribbon and abeliansubcategories of repsH whose simple objects are scalar, and ϕB is the morphism(3.10) corresponding to B. By Corollary 3.9, we have that

I(H)s ⊂ I(H).

Note that this inclusion may be strict (see Remark 4.12).Let V be a set of representatives of isomorphism classes of indecomposable finite-

dimensional left H-modules with non-zero quantum dimension. Note that π(V) ={π(V ) |V ∈ V} is a set of representatives of isomorphism classes of simple objects ofrepsH . Let λ be a non-zero right integral for H∗. Since H∗ is a free right H-modulewith basis λ (see Section 4.1), there exists a (unique) element zV ∈ H such that

(4.36) λ(zV x) = Tr(G−1x idV )

for all x ∈ H , where G is the special grouplike element of H . Recall that dimq(V ) =Tr(G−1idV ) denotes the quantum dimension of V (see Lemma 4.2).

Corrolary 4.10. (a) If z ∈ I(H)s, then z = k∑

V ∈W dimq(V ) zV for somefinite subset W of V and some scalar k ∈ k.

(b) Let W be a (finite) set of representatives of isomorphism classes of simpleobjects of a full ribbon and abelian subcategory of repsH . We can supposethat W ⊂ π(V). If the objects of W are scalar, then

∑

V ∈π−1(W)

dimq(V ) zV ∈ I(H)s.

Proof. Let B be finitely semisimple full ribbon and abelian subcategory of repsHwhose simple objects are scalar, and let (B, j) be the coend of the functor (1.16)of B (as in Section 3.2). We can suppose that there exists a (finite) subset Wof V such that π(W) is a set of representatives of isomorphism classes of simpleobjects of B. Recall that B = ⊕V ∈W π(V )∗ ⊗ π(V ). In particular, there existmorphisms pV : B → π(V )∗ ⊗ π(V ) and qV : π(V )∗ ⊗ π(V ) → B of B such thatidB =

∑V ∈W qV pV and pV qW = δV,W idπ(V )∗⊗π(V ). Recall that jV = qV for any

V ∈ W . Let φ : L(H) → HomrepH(k, A) be as is Section 4.6. As in (3.10), we set

ϕB =∑

V ∈W

π(iV )pV ∈ HomrepsH(B, π(A)).

Let V ∈ W . Let (ei)i be a basis of V with dual basis (e∗i )i. By Lemma 4.2(b) and(4.25) we have that

〈iV coevV (1k), x〉 = 〈iV (idV ∗ ⊗G−1idV )σV,V ∗ coevV (1k), x〉

=∑

i〈iV (e∗i ⊗G−1ei), x〉 =

∑i〈e

∗i , xG

−1ei〉

= Tr(xG−1idV ) = Tr(G−1x idV )

= λ(zV x) = 〈φzV (1k), x〉,

for any x ∈ H , that is, iV coevV = φzV . Moreover,

ϕBjπ(V )coevπ(V ) =∑

W∈W

π(iW )pW jπ(V )coevπ(V )

=∑

W∈W

δV,W π(iW )π(coevV )

= π(iV coevV ) = π(φzV ).


Hence Part (a) follows from Lemma 3.3 and Corollary 3.9(a), and Part (b) followsfrom Theorem 3.4 and Corollary 3.9(a). �

Lemma 4.11. If H is not semisimple, then ε(z) = 0 for any z ∈ I(H)s.

Remark 4.12. When H is not semisimple, it is possible that I(H)s I(H). Forexample, if H is unimodular but not semisimple, then 1 ∈ I(H) (by Corollary 4.9)and 1 6∈ I(H)s (by Lemma 4.11, since ε(1) = 1).

Proof. Let Λ be a left integral for H such that λ(Λ) = 1. Since H is not semisimple,we have ε(Λ) = 0 (by [Abe80, Theorem 3.3.2]) and Λ2 = ε(Λ)Λ = 0. Now, ifM is a finite-dimensional left H-module, then (Λ idM )2 = Λ2idM = 0 and soTr(Λ idM ) = 0. Let z ∈ I(H)s. By Corollary 4.10(a), there exist k ∈ k and a finitesubset W of V such that z = k

∑V ∈W dimq(V ) zV . Then

λ(zΛ) = k∑

V ∈W

dimq(V )Tr(G−1Λ idV ) = k∑

V ∈W

dimq(V )ε(G−1)Tr(Λ idV ) = 0.

Hence ε(z) = ε(z)λ(Λ) = λ(ε(z)Λ) = λ(zΛ) = 0. �

Recall (see [CR62]) that k is a splitting field for a k-algebra A if every simplefinite dimensional left A-module is scalar. Note that this is always the case if k isalgebraically closed.

Corrolary 4.13. If H is semisimple and k is a splitting field for H, then I(H)s =I(H) and this set is composed by elements z ∈ Z(H) satisfying S(z) = z andλ(zx(1))zx(2) = λ(zx)z for all x ∈ H.

Proof. SinceH is semisimple and finite-dimensional, we have that repsH = repH andthat V finite. Moreover, since k is a splitting field for a H , every V ∈ V is scalar.Then I(H) = φ−1(I(repH)) ⊂ I(H)s and so I(H)s = I(H). Moreover, since H isunimodular (because it is semisimple), we have that L(H) = Z(H), N(H) = 0, andT (x) = S(x) for all x ∈ H (see Section 4.6). Therefore, by using Theorem 4.8, weget that z ∈ I(H) if and only if z ∈ Z(H), S(z) = z, and λ(zx(1))zx(2) = λ(zx)zfor all x ∈ H . �

Proposition 4.14. Suppose that H is semisimple, that k is of characteristic 0, andthat k is a splitting field for H. Then Z(H) =

⊕V ∈V kzV and

∑V ∈V dimq(V ) zV ∈

k∗1.

Proof. Note that H is cosemisimple since any finite-dimensional semisimple Hopfalgebra over a field of characteristic 0 is cosemisimple (see [LR88, Theorem 3.3]).Then S2 = idH by [LR87, Theorem 4] and λ(1) 6= 0 by [Abe80, Theorem 3.3.2].

Note that V is finite (since H is finite-dimensional). By [CR62, Theorem 25.10],any simple left ideal of H is isomorphic (as a left H-module) to a (unique) elementof V . For any V ∈ V , let HV ⊂ H be the sum of all the simple left ideals of H whichare isomorphic to V . By [CR62, Theorem 25.15], HV is a two-sided ideal of H ,HV is simple k-algebra (the operations being those induced by H), BVBW = 0 forV 6= W ∈ V , H =

⊕V ∈V HV , and H is isomorphic (as an algebra) to

∏V ∈V HV .

Moreover, if eV denotes the unit of HV , then 1 =∑

V ∈V eV , HV = HeV , andeV eW = δV,W eV for all V,W ∈ V . By [CR62, Theorem 26.4], since HV is a simplek-algebra, V is a simple left HV -module, and EndHV

(V ) = k (because V is a scalarH-module), we have that HV is isomorphic (as an algebra) to Endk(V ) and thatdimk(V ) is the number of simple left ideals appearing in a direct sum decomposition


of HV as such a sum. Then Z(HV ) = keV (since Z(Endk(V )) = kidV ) and soZ(H) =

⊕V ∈V keV . Moreover, for any x ∈ H , we have that

(4.37) Tr(x idH) =∑

V ∈V

Tr(x idHV) =

∑

V ∈V

dimk(V )Tr(x idV ).

The map x ∈ H 7→ Tr((x idH) ◦ S2) ∈ k is a right integral for H∗ (by [Rad94b,Proposition 2(b)] applied to Hop). Therefore, since S2 = idH and by the unique-ness of integrals, there exists k ∈ k such that Tr(x idH) = k λ(x) for all x ∈ H .Evaluating with x = 1, we get that k = dimk(H)/λ(1). Hence, by (4.37),

(4.38) k λ(x) =∑

V ∈V

dimk(V )Tr(x idV )

for all x ∈ H .Let V ∈ V . By (4.21) and since S2 = idH , the special grouplike element G

of H is central and so G−1idV is H-linear. Therefore, since V is scalar and G isinvertible, there exists a (unique) γV ∈ k∗ such that G−1idV = γV idV . Since H andH∗ are semisimple and so unimodular, their special grouplike elements are trivial.Then G2 = 1 by (4.22) and so γ2V = 1. Hence, for all x ∈ H ,

dimq(V )Tr(G−1x idV ) = Tr(G−1idV )Tr(xG−1idV )

= γ2V Tr(idV )Tr(x idV ) = dimk(V )Tr(x idV ).(4.39)

For any x ∈ H , we have that:

λ(dimq(V ) zV x) = dimq(V )Tr(G−1x idV ) by (4.36)

= dimk(V )Tr(x idV ) by (4.39)

=∑

W∈V dimk(W )Tr(xeV idW ) since eV idW = δV,W idV

= λ(k eV x) by (4.38),

and so dimq(V ) zV = k eV (since H∗ is a free right H-module with basis λ).Finally, since k = dimk(H)/λ(1) 6= 0 (because the characteristic of k is 0) and

dimq(V ) 6= 0 (because repH is semisimple, see Section 3.1), we get that Z(H) =⊕V ∈V keV =

⊕V ∈V kzV and

∑V ∈V dimq(V ) zV = k

∑V ∈V eV = k 1 ∈ k∗1. �

4.8. HKR-type invariants. Let H be a finite-dimensional ribbon Hopf algebra.We use notations of Section 4.6.

By Theorem 2.5 and Proposition 2.3, for any z ∈ I(H)norm,

(4.40) τ(H,z)(M) = τrepH(M ;φz) ∈ k

is an invariant of 3-manifolds. Note that the choice of the normalisation in the defi-nition of τrepH

(M ;φz) (see Proposition 2.3) implies that τ(H,z)(M) does not dependon the choice of the non-zero right integral λ for H∗ used to define τrepH

(M ;φz).By Remarks 2.6, for any z ∈ I(H)norm, we have that

τ(H,z)(S3) = 1, τ(H,z)(M#M ′) = τ(H,z)(M) τ(H,z)(M

′),

for all 3-manifoldsM , M ′. Moreover, if z ∈ I(H)norm, n ∈ N(H), and k ∈ k∗, thenkz + n ∈ I(H)norm and, for all 3-manifold M ,

(4.41) τ(H,kz+n)(M) = τ(H,z)(M).


Definition 4.15. An invariant of closed 3-manifolds I with values in k is said tobe of HKR-type if there exist a finite-dimensional ribbon Hopf algebra H (over k)and an element z ∈ I(H)norm such that I(M) = τ(H,z)(M) for all 3-manifolds M .

In Proposition 4.18, we show that the Reshetikhin-Turaev invariants defined withpremodular Hopf algebras are of HKR-type.

Let us show that any HKR-type invariant can be computed by using the Kauffman-Radford algorithm (which is given in [KR95] for the case H unimodular and forz = 1). Fix a finite-dimensional ribbon Hopf algebra H , a non-zero right integral λfor H∗, and an element z ∈ I(H)norm. LetM be a 3-manifold and L = L1∪· · ·∪Ln

be a framed link in S3 such that M ≃ ML. Let us recall the Kauffman-Radfordalgorithm1:

(A). Consider a diagram D of L (with blackboard framing). Each crossing ofD is decorated with the R-matrix R =

∑i ai ⊗ bi as in Figure 11. The diagram

obtained after this step is called the flat diagram of D. Note that the flat diagramof D is composed by n closed plane curves, each of them arising from a componentof L.

PSfrag replacements

aibi

bi

∑

i

∑

i

S(ai)

Figure 11.

(B). On each component of the flat diagram of D, the algebraic decoration isconcentrated in an arbitrary point (other than extrema and crossings) according tothe rules of Figure 12, where a, b ∈ H .

PSfrag replacements

a

a aaa

a S(a) aab ab 1S(a)

G−1G

Figure 12.

In that way we get an element∑

k vk1 ⊗ · · · ⊗ vkn ∈ H⊗n, where vki corresponds to

the component of the flat diagram of D arising from Li, see Figure 13.

PSfrag replacements

vk1 vk2

vkn

a

L1 L2 Ln

. . . . . .∑

k

Figure 13.

1This corresponds to [KR95] applied with the opposite ribbon Hopf algebra Hop to H.


For 1 ≤ i ≤ n, let di be the Whitney degree of the flat diagram of Li obtainedby traversing it upwards from the point where the algebraic decoration have beenconcentrated. The Whitney degree is the total turn of the tangent vector to thecurve when one traverses it in the given direction, see Figure 14.

PSfrag replacements

d = 1 d = −1 di = −2vki

Figure 14.

Proposition 4.16.

τ(H,z)(M) = λ(zθ)b−(L)−nL λ(zθ−1)−b−(L)∑

k

λ(zGd1+1vk1 ) · · ·λ(zGdn+1vkn).

Proof. Choose an orientation for L. Let T be a n-special tangle such that L is iso-topic T ◦(∪−⊗· · ·⊗∪−), where the ith cup (with clockwise orientation) correspondsto the component Li. Let DT be a diagram of T . Note that D = DT ◦ (∪⊗ · · ·⊗∪)is a diagram of L. Apply steps (A) and (B) to DT , see Figure 15. Note that, inthis case, di = −1.

PSfrag replacements

vk1 vkn. . .

T

∑

k

Figure 15.

From the definition of the monoidal structure, duality, braiding and twist ofrepH (see Section 4.4), it is not difficult to verify that, for any finite-dimensionalleft H-modules M1, . . . ,Mn,

T(M1,...,Mn) =∑

k

evM1(idM∗1⊗ v1k idM1)⊗ · · · ⊗ evMn

(idM∗n⊗ vnk idMn

).

Then, by Lemma 4.3,

τrepH(L;φz) = φT ◦ φ⊗n

z =∑

k

evH(φz(1k)⊗ v1k)⊗ · · · ⊗ evH(φz(1k)⊗ vnk )

=∑

k

λ(zvk1 ) · · ·λ(zvkn) =

∑

k

λ(zGd1+1vk1 ) · · ·λ(zGdn+1vkn).

Hence we get the results since Θ±φz = λ(zθ±1). �

Corrolary 4.17. Suppose that H is unimodular and λ(θ) 6= 0 6= λ(θ−1). Then1 ∈ I(H)norm and τ(H,1)(M) is the Hennings-Kauffman-Radford invariant of 3-manifolds defined with the opposite ribbon Hopf algebra Hop to H.

Proof. This is an immediate consequence of Corollary 4.9, Proposition 4.16, and thedefinition of the Hennings-Kauffman-Radford invariant given, e.g., in [KR95]. �


4.9. Reshetikhin-Turaev from premodular Hopf algebras. Let (H,V) be afinite-dimensional premodular Hopf algebra. This means (see [Tur94]) that H isa finite-dimensional ribbon Hopf algebra and V is a finite set of finite-dimensionalpairwise non-isomorphic left H-modules such that:

• each V ∈ V is non-negligible scalar;• the trivial left H-module k belongs to V ;• for any V ∈ V , there exists W ∈ V such that V ∗ ≃W ;• for any V,W ∈ V , V ⊗W splits as a (finite) direct sum of certain modulesof V (possibly with multiplicities) and a negligible H-module.

By negligible H-module, we mean a finite-dimensional left H-module N such thattrq(f) = 0 for any f ∈ EndrepH

(N) or, equivalently, such that dimq(N) = 0.Consider the semisimplification repsH of repH (see Section 3.4) and let π : repH →

repsH be the ribbon functor associated to this semisimplification.Let BV be the full subcategory of repsH whose objects are finite directs sums of

objects of π(V) = {π(V ) |V ∈ V}. By the definition of a premodular Hopf algebra,we have that BV is a full ribbon subcategory of repsH . Note that BV is a finitelysemisimple ribbon k-category whose simples objects are scalar and has π(V) as a(finite) set of representatives of isomorphism classes of simple objects. Recall thatthe Reshetikhin-Turaev invariant RTBV

(M) of 3-manifolds is well-defined when

∆BV

± 6= 0 (see Section 3.3).Let λ be a non-zero right integral for H∗. For any V ∈ V , as in (4.36), we let

zV ∈ H such that λ(zV x) = Tr(G−1x idV ) for all x ∈ H . Set

(4.42) zV =∑

V ∈V

dimq(V ) zV ,

where dimq(V ) = Tr(G−1idV ) is the quantum dimension of V . By Corollary 4.10(b),we have that zV ∈ I(H).

Proposition 4.18. If ∆BV

± 6= 0, then zV ∈ I(H)norm and τ(H,zV)(M) = RTBV(M)

for all 3-manifold M .

Note that Proposition 4.18 says that the Reshetikhin-Turaev invariant definedfrom a premodular Hopf algebra is of HKR-type.

Proof. Let (A, i) be the coend of the functor (1.16) of repH (as in Section 4.5),let (B, j) be the coend of the functor (1.16) of BV (as in Section 3.2). Set αBV

=∑V ∈V dimq(V )jπ(V )coevπ(V ). Suppose that ∆BV

± 6= 0. By Corollary 3.7, we havethat αBV

∈ I(BV)norm and RTBV

(M) = τBV(M ;αBV

) for all 3-manifold M . SetϕBV

: B → π(A) as in (3.10). As in the proof of Corollary 4.10, we have thatπ(φzV ) = ϕBV

jπ(V )coevπ(V ). Then π(φzV ) = ϕBVαBV

. Since αBV∈ I(BV)

norm andπ(φzV ) = ϕαBV

, Corollary 3.9(b) gives that φzV ∈ I(repH)norm and τBV(M ;αBV

) =τrepH

(M ;φzV ) for all 3-manifold M . Hence zV ∈ I(H)norm and τ(H,zV)(M) =RTBV

(M) for all 3-manifold M . �

Note that if H is semisimple finite-dimensional ribbon Hopf algebra, k is a split-ting field for H , and V is a set of representatives of isomorphism classes of simpleleft H-modules, then (H,V) is a premodular Hopf algebra and BV = repH .

Corrolary 4.19. Let H be a finite-dimensional semisimple ribbon Hopf algebra.Suppose that the base field k is of characteristic 0 and is a splitting field for H. Thenthe Hennings-Kauffman-Radford invariant of 3-manifolds computed with Hop and


the Reshetikhin-Turaev invariant of 3-manifolds computed with repH are simulta-neously well-defined (that is, ∆

repH

± 6= 0 if and only if 1 ∈ I(H)norm). Moreover,if they are, then they coincide, that is, τ(H,1)(M) = RTrepH

(M) for all 3-manifoldM .

Remark 4.20. Conclusions of Corollary 4.19 may be no more true when H isnot semisimple (see Remark 4.12). Moreover, in the modular case (in the sense ofRemark 3.5), Corollary 4.19 was first shown in [Ker97].

Proof. By Proposition 4.18, the Reshetikhin-Turaev invariant of 3-manifolds com-puted from repH is well-defined if zV =

∑V ∈V dimq(V ) zV ∈ I(H)norm and is

equal to τ(H,zV). By Corollary 4.17, the Hennings-Kauffman-Radford invariant of3-manifolds computed with Hop is well-defined if 1 ∈ I(H)norm and is equal toτ(H,1). Now, by Proposition 4.14,

∑V ∈V dimq(V ) zV = k 1 for some k ∈ k∗. We

conclude by using (4.41). �

5. A non-unimodular example

Let us examine the case of family of non-unimodular ribbon Hopf algebras, de-fined by Radford [Rad94a], which includes Sweedler’s Hopf algebra.

Let n be an odd positive integer and k be a field whose characteristic does notdivide 2n. LetHn be the k-algebra generated by a and x with the following relations

a2n = 1, x2 = 0, ax = −xa.

The algebra Hn is a Hopf algebra for the following structure maps:

∆(a) = a⊗ a, ∆(x) = x⊗ an + 1⊗ x,

ε(a) = 1, ε(x) = 0,

S(a) = a−1, S(x) = anx.

The set B = {alxm | 0 ≤ l < 2n, 0 ≤ m ≤ 1} is a basis for Hn. The dual basis of B

is {akxr | 0 ≤ k < 2n, 0 ≤ r ≤ 1}, where akxn(alxr) = δl,k δm,r. Set

Λ = (1 + a+ a2 + · · ·+ a2n−1)x and λ = anx.

Then Λ is a left integral for Hn and λ is a right integral for H∗n such that λ(Λ) =

λ(S(Λ)) = 1. The distinguished grouplike element of Hn is g = an ∈ G(Hn) andthe distinguished grouplike element ν ∈ G(H∗

n) = Alg(Hn, k) of H∗n is given by

ν(a) = −1 and ν(x) = 0.Suppose that k has a primitive 2n-root of unity ω. Let an odd integer 1 ≤ s < 2n

and β ∈ k. Then

Rω,s,β =1

2n

∑

0≤i,l<2n

w−il ai ⊗ asl +β

2n

∑

0≤i,l<2n

w−il aix⊗ asl+nx

is a R-matrix for Hn and hν = (idHn⊗ ν)(Rω,s,β) = an.

The set {el =12n

∑0≤i<2n ω

−ilai | 0 ≤ l < 2n} is a basis of the algebra k[a] andverifies:

eiej = δi,j ei, 1 = e0 + · · ·+ e2n−1, aiej = ωijej ,

ε(el) = δl,0, ∆(el) =∑

0≤i<2n

ei ⊗ el−i, S(el) = e−l.


Denote by χ : k∗ → k[a] the algebra homomorphism defined by

α ∈ k∗ 7→ χ(α) =∑

0≤l<2n

αl2 el.

The quasitriangular Hopf algebra (Hn, Rω,s,β) is ribbon with twist θ = anχ(ωs).Note that θp = apnχ(ωps) for any integer p. The special grouplike element of Hn

is then G =∑

0≤l<2n(−1)l el = an.

Let T : Hn → Hn, L(Hn), N(Hn), V2(Hn), I(Hn), and I(Hn)norm be as in

Section 4.6. It is not difficult to verify that

T (ak) = (−1)kan−k and T (akx) = a−kx for all 0 ≤ k < 2n,

L(Hn) = kx⊕ kax⊕ · · · ⊕ ka2n−1x,

Z(Hn) = k1⊕ ka2 ⊕ · · · ⊕ ka2n−2,

N(Hn) = kx⊕ ka2x⊕ · · · ⊕ ka2n−2x = Z(Hn)x,

V2(Hn) = ⊕0≤p,q<n

k(a2p ⊗ a2q)⊕ ⊕0≤k,l<2n

k(akx⊗ alx).

Let z ∈ L(Hn). Since a2n = 1, we can write z =∑

k∈Z/2nZ αkakx for some

function α : Z/2nZ→ k. Using Theorem 4.8, we get that z ∈ I(Hn) if and only if

α−k = αk and αkαk+l−n = αkαl.

for all k, l odd. Let γ : α : Z/nZ → k defined by γk = α2k+n. Since n is odd, wehave that z −

∑k∈Z/nZ γka

2k+nx ∈ N(Hn). Then z ∈ I(Hn) if and only if

(5.1) γ−k = γk and γkγk+l = γkγl.

for all k, l.Suppose that z ∈ I(Hn) and z 6= 0. Using (5.1), we get that γkγ0 = γkγ−k = γ2k

for all k. In particular, γ0 6= 0 and γk = γ0 whenever γk 6= 0. Set

d = min{1 ≤ k ≤ n | γk 6= 0} ≥ 1.

Note that γk = 0 for all 1 ≤ k < d and, by (5.1), that γk+d = γk for all k. Theinteger d divides n. Indeed, let r such that rd ≤ n < rd+ d. Then 0 ≤ n− rd ≤ nand γn−rd = γn = γ0 6= 0. Therefore, by definition of d, we get that n − rd = 0and so d|n. Hence z = γ0zd + w, where w ∈ N(Hn) and

zd =

nd−1∑

k=0

a2dk+nx.

Conversely, it is not difficult to verify that zd ∈ I(Hn) whenever d divides n.Hence

I(Hn) =⋃

d|n

{αzd + w |α ∈ k and w ∈ N(Hn)}.

Let d be an positive integer dividing n. For any α ∈ k and w ∈ N(Hn), we havethat

λ((αzd + w)θ±1) =α

2d

2d−1∑

k=0

(ω

nd

)±sndk2+nk

.

The sum of the right-hand sum of this equality is a Gauss sum which is non-zero ifand only if the enhancement k ∈ Z/2dZ 7→ ψ(k) = ±snd k

2 + nk ∈ Z/2dZ is tame,that is, ψ(x) = 0 for any x ∈ Z/2dZ such that ψ(x + y) = ψ(x) + ψ(y) for all


y ∈ Z/2dZ, see [Tay]. Since n and s are odd, it is not difficult to verify that ψ istame. Therefore λ((αzd + w)θ±1) 6= 0 if and only if α 6= 0. Hence

I(Hn)norm =

⋃

d|n

{αzd + w |α ∈ k∗ and w ∈ N(Hn)}.

In conclusion, the ribbon Hopf algebra Hn leads to D(n) HKR-type invariantsof 3-manifolds, where D(n) denotes the number of positive divisors of n, which areτ(Hn,zd) where 1 ≤ d ≤ n and d|n.

Note that Hn is not unimodular (and so neither semisimple) since ν 6= ε. There-fore 1 6∈ I(Hn) (by Corollary 4.9), that is, the Hennings-Kauffman-Radford invari-ant is not defined for Hn. Moreover, the categorical Hopf algebra A = H∗

n of repHn

does not possess any non-zero two-sided integral (since Hn is not unimodular), andso the Lyubashenko invariant of 3-manifolds is not defined for repHn

.

Appendix A. Traces on ribbon Hopf algebras

Recall that a trace on a Hopf algebra H is a form t ∈ H∗ such that t(xy) = t(yx)and t(S(x)) = t(x) for all x, y ∈ H .

Let H be a finite-dimensional ribbon Hopf algebra. Let λ ∈ H∗ be a non-zeroright integral for H∗, ν ∈ G(H∗) = Alg(H, k) be the distinguished grouplike ele-ment of H∗ and G be the special grouplike element of H . Recall that · denotes theright action of H on H∗ defined in (4.1), that ↼ denotes the right H∗-action on Hdefined in (4.2), that L(H) denotes the k-subspace ofH constituted by the elementsz ∈ H verifying (x ↼ ν)z = zx for all x ∈ H , and that T denotes the k-endomor-phism ofH defined by z 7→ T (z) = (S(z)↼ ν)hν , where hν = (idH⊗ν)(R) ∈ G(H).

In the next proposition, we give an algebraic description of the space of traceson H .

Proposition A.1. The space {z ∈ L(H) |T (z) = z} is k-isomorphic to the spaceof traces on H via the map z 7→ λ · (zG).

If H is unimodular, then L(H) = Z(H) and T = S, and so we recover theparameterization of traces on H given in [Rad94b, Hen96].

Proof. Let z ∈ L(H) such that T (z) = z. Set t = λ · (zG) ∈ H∗. By using (4.26)and (4.21), we have that for any x, y ∈ H ,

t(xy) = λ(zGxy) = λ(zS2(y)Gx) = λ(zGyx) = t(yx).

Moreover, for any x ∈ H ,

t(S(x)) = λ(zGS(x)) = λ(G2hνxG−1S−1(z)) by (4.23)

= λ((S(z)↼ ν)G2hνxG−1) by (4.3)

= λ(zh−1ν G2hνxG

−1) since T (z) = z

= λ(zS2(G−1)h−1ν G2hνx) by (4.26)

= λ(zG−1h−1ν G2hνx)

= λ(zGS−4(h−1ν )hνx) by (4.21)

= λ(zGh−1ν hνx) = λ(zGx) = t(x).


Conversely, let t ∈ H∗ be a trace on H . Since (H∗, ·) is a free right H-moduleof rank 1 with basis λ and G is invertible, there exists a (unique) z ∈ H such thatt = λ · (zG). Let x ∈ H . For all a ∈ H ,

λ(zxGy) = λ(zGS−2(x)y) by (4.21)

= t(S−2(x)y) = t(yS−2(x))

= λ(zGyS−2(x))

= λ((x ↼ ν)zGy) by (4.3).

Therefore, since (H∗, ·) is a free right H-module of rank 1 with basis λ, we havethat (x ↼ ν)zG = zxG and so (x ↼ ν)z = zx. Hence z ∈ L(H). Moreover, for allx ∈ H ,

λ(zGx) = t(x) = t(S(x)) = λ(zGS(x))

= λ(G2hνxG−1S−1(z)) by (4.23)

= λ(G2hνxS−1(zG))

= λ(G2hνxS−1((G ↼ ν)z)) since z ∈ L(H)

= λ((S((G ↼ ν)z)↼ ν)G2hνx) by (4.3).

Therefore, since (H∗, ·) is a free right H-module of rank 1 with basis λ,

zG = (S((G ↼ ν)z)↼ ν)G2hν

= (S(z)↼ ν)(S(G ↼ ν)↼ ν)GS2(hν)G by (4.21)

= (S(z)↼ ν)ν(G)ν(G−1)G−1GhνG

= (S(z)↼ ν)hνG = T (z)G

and so T (z) = z. �

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Departement des Sciences Mathematiques, Universite Montpellier II, Case Courrier

051, Place Eugene Bataillon, 34095 Montpellier Cedex 5, France

E-mail address: [email protected]

kirby elements and quantum invariants …2 alexis virelizier the kirby theorem [kir78], we have that...

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